CHAPTER FOURTEEN ACIDS AND BASES...346 CHAPTER 14 ACIDS AND BASES bases. The nitrogen atom has an...

345

CHAPTER FOURTEEN

ACIDS AND BASES

Questions

16. The Arrhenius definitions are: acids produce H in water and bases produce OH in water. The+ -

difference between strong and weak acids and bases is the amount of H and OH produced.+ -

a. A strong acid is 100% dissociated in water.b. A strong base is 100% dissociated in water.c. A weak acid is much less than 100% dissociated in water (typically 1-10% dissociated).d. A weak base is one that has only a small percentage of molecules react with water to form OH .-

17. a. Arrhenius acid: produce H in water+

b. Brønsted-Lowry acid: proton (H ) donor+

c. Lewis acid: electron pair acceptor

The Lewis definition is most general. The Lewis definition can apply to all Arrhenius and Brønsted-Lowry acids; H has an empty 1s orbital and forms bonds to all bases by accepting a pair+

of electrons from the base. In addition, the Lewis definition incorporates other reactions not

3 3 3 3 3typically considered acid-base reactions, e.g., BF (g) + NH (g) ÷ F BSNH (s). NH is something we

3usually consider a base and it is a base in this reaction using the Lewis definition; NH donates a pairof electrons to form the NSB bond.

2 318. When a strong acid (HX) is added to water, the reaction HX + H O ÷ H O + X basically goes to+ -

3completion. All strong acids in water are completely converted into H O and X . Thus, no acid+ -

3stronger than H O will remain undissociated in water. Similarly, when a strong base (B) is added to+

2water, the reaction B + H O ÷ BH + OH basically goes to completion. All bases stronger than OH+ - -

3are completely converted into OH and BH . Even though there are acids and bases stronger than H O- + +

3and OH , in water these acids and bases are completely converted into H O and OH .- + -

2 w19. H O º H + OH K = [H ] [OH ] = 1.0 × 10+ - + - -14

Neutral solution: [H ] = [OH ]; [H ] = 1.0 × 10 M; pH = -log (1.0 × 10 ) = 7.00+ - + -7 -7

20. 10.78 (4 S.F.); 6.78 (3 S.F.); 0.78 (2 S.F.); A pH value is a logarithm. The numbers to the left of thedecimal place identy the power of ten to which [H ] is expressed in scientific notation, e.g.,+

10 , 10 , 10 . The number of decimal places in a pH value identifies the number of significant-11 -7 -1

figures in [H ]. In all three pH values, the [H ] should be expressed only to two significant figures+ +

since these pH values have only two decimal places.

21. Neutrally charged organic compounds containing at least one nitrogen atom generally behave as weak

CHAPTER 14 ACIDS AND BASES346

bases. The nitrogen atom has an unshared pair of electrons around it. This lone pair of electrons isused to form a bond to H .+

22. a. The weaker the X S H bond, the stronger the acid.b. As the electronegativity of neighboring atoms increases, the strength of the acid increases.c. As the number of oxygen atoms increases, the strength of the acid increases.

23. In general, the weaker the acid, the stronger the conjugate base and vice versa.

a. Since acid strength increases as the X S H bond strength decreases, conjugate base strength willincrease as the strength of the X S H bond increases.

b. Since acid strength increases as the electronegativity of neighboring atoms increases, conjugatebase strength will decrease as the electronegativity of neighboring atoms increases.

c. Since acid strength increases as the number of oxygen atoms increases, conjugate base strengthdecreases as the number of oxygen atoms increase.

24. A Lewis acid must have an empty orbital to accept an electron pair, and a Lewis base must have anunshared pair of electrons.

2 3 225. a. H O and CH CO -

b. An acid-base reaction can be thought of as a competition between two opposing bases. Since this

a 3 2 2equilibrium lies far to the left (K < 1), CH CO is a stronger base than H O.-

c. The acetate ion is a better base than water and produces basic solutions in water. When we putacetate ion into solution as the only major basic species, the reaction is:

3 2 2 3 2CH CO + H O º CH CO H + OH- -

3 2Now the competition is between CH CO and OH for the proton. Hydroxide ion is the strongest- -

bbase possible in water. The equilibrium above lies far to the left, resulting in a K value less than

b 2one. Those species we specifically call weak bases (10 < K < 1) lie between H O and OH in-14 -

base strength. Weak bases are stronger bases than water but are weaker bases than OH .-

4 2 326. The NH ion is a weak acid because it lies between H O and H O (H ) in terms of acid strength.+ + +

Weak acids are better acids than water, thus their aqueous solutions are acidic. They are weak acids

3because they are not as strong as H O (H ). Weak acids only partially dissociate in water and have+ +

aK values between 10 and 1.-14

27. When an acid dissociates, ions are produced. The conductivity of the solution is a measure of thenumber of ions. In addition, the colligative properties, which are discussed in Chapter 11, depend onthe number of particles present. So measurements of osmotic pressure, vapor pressure lowering,freezing point depression or boiling point elevation will also allow us to determine the extent ofionization.

a b w a b w28. K × K = K , -log (K × K ) = -log K

CHAPTER 14 ACIDS AND BASES 347

a b w a b w-log K - log K = -log K , pK + pK = pK = 14.00 (at 25°C)

Exercises

Nature of Acids and Bases

4 2 3 4 429. a. HClO (aq) + H O(l) ÷ H O (aq) + ClO (aq). Only the forward reaction is indicated since HClO+ -

is a strong acid and is basically 100% dissociated in water. For acids, the dissociation reactionis commonly written without water as a reactant. The common abbreviation for this reaction is:

4 4 aHClO (aq) ÷ H (aq) + ClO (aq). This reaction is also called the K reaction as the equilibrium+ -

aconstant for this reaction is called K .

b. Propanoic acid is a weak acid, so it is only partially dissociated in water. The dissociation

3 2 2 2 3 3 2 2reaction is: CH CH CO H(aq) + H O(l) H O (aq) + CH CH CO (aq) or+ -

3 2 2 3 2 2CH CH CO H(aq) H (aq) + CH CH CO (aq).+ -

4c. NH is a weak acid. Similar to propanoic acid, the dissociation reaction is:+

4 2 3 3 4 3NH (aq) + H O(l) H O (aq) + NH (aq) or NH (aq) H (aq) + NH (aq)+ + + +

a30. The dissociation reaction (the K reaction) of an acid in water commonly omits water as a reactant.We will follow this practice. All dissociation reactions produce H and the conjugate base of the acid+

that is dissociated.

2 3 2 2 3 2 aa. HC H O (aq) H (aq) + C H O (aq) K = + -

2 6 2 5 ab. Co(H O) (aq) H (aq) + Co(H O) (OH) (aq) K = 3+ + 2+

3 3 3 2 ac. CH NH (aq) H (aq) + CH NH (aq) K = + +

31. An acid is a proton (H ) donor and a base is a proton acceptor. A conjugate acid-base pair differs by+

only a proton (H ).+

Conjugate ConjugateAcid Base Base of Acid Acid of Base

2 3a. HF H O F H O- +

2 4 2 4 3b. H SO H O HSO H O- +

4 2 4 3c. HSO H O SO H O- 2- +

32. Conjugate ConjugateAcid Base Base of Acid Acid of Base


2 6 2 2 5 3a. Al(H O) H O Al(H O) (OH) H O3+ 2+ +

3 2 2 3b. HONH H O HONH H O+ +

6 5 2 6 5 3c. HOCl C H NH OCl C H NH- +

a a a33. Strong acids have a K >> 1 and weak acids have K < 1. Table 14.2 in the text lists some K values

afor weak acids. K values for strong acids are hard to determine so they are not listed in the text.However, there are only a few common strong acids so if you memorize the strong acids, then all other

3 4 2 4acids will be weak acids. The strong acids to memorize are HCl, HBr, HI, HNO , HClO and H SO .

4a. HClO is a strong acid.

ab. HOCl is a weak acid (K = 3.5 × 10 ).-8

2 4c. H SO is a strong acid.

2 3 a1 a2d. H SO is a weak diprotic acid since the K and K values are less than one.

34. The beaker on the left represents a strong acid in solution; the acid, HA, is 100% dissociated into theH and A ions. The beaker on the right represents a weak acid in solution; only a little bit of the acid,+ -

HB, dissociates into ions, so the acid exists mostly as undissociated HB molecules in water.

2a. HNO : weak acid beaker

3b. HNO : strong acid beakerc. HCl: strong acid beakerd. HF: weak acid beaker

2 3 2e. HC H O : weak acid beaker

a a35. The K value is directly related to acid strength. As K increases, acid strength increases. For water,

w ause K when comparing the acid strength of water to other species. The K values are:

3 a aHNO : strong acid (K >> 1); HOCl: K = 3.5 × 10-8

4 a 2 a wNH : K = 5.6 × 10 ; H O: K = K = 1.0 × 10+ -10 -14

a 3 4 2From the K values, the ordering is: HNO > HOCl > NH > H O. +

36. Except for water, these are the conjugate bases of the acids in the previous exercise. In general, the

3weaker the acid, the stronger the conjugate base. NO is the conjugate base of a strong acid; it is a-

3 2 3terrible base (worse than water). The ordering is: NH > OCl > H O > NO - -

a w37. a. HCl is a strong acid and water is a very weak acid with K = K = 1.0 × 10 . HCl is a much-14

2stronger acid than H O.

2 a w 2 a 2 2 ab. H O, K = K = 1.0 × 10 ; HNO , K = 4.0 × 10 ; HNO is a stronger acid than H O since K-14 -4

2 a 2for HNO > K for H O.

6 5 a a 6 5c. HOC H , K = 1.6 × 10 ; HCN, K = 6.2 × 10 ; HCN is a stronger acid than HOC H since-10 -10

a a 6 5K for HCN > K for HOC H .

2 b38. a. H O; The conjugate bases of strong acids are terrible bases (K < 10 ). -14


2 bb. NO ; The conjugate bases of weak acids are weak bases (10 < K < 1).- -14

6 5 a b wc. OC H ; For a conjugate acid-base pair, K × K = K . From this relationship, the stronger the-

b aacid the weaker the conjugate base (K decreases as K increases). Since HCN is a stronger acid

6 5 a a 6 5 6 5than HOC H (K for HCN > K for HOC H ), OC H will be a stronger base than CN .- -

Autoionization of Water and the pH Scale

w39. At 25°C, the relationship: [H ] [OH ] = K = 1.0 × 10 always holds for aqueous solutions. When+ - -14

[H ] is greater than 1.0 × 10 M, the solution is acidic; when [H ] is less than 1.0 × 10 M, the+ -7 + -7

solution is basic; when [H ] = 1.0 × 10 M, the solution is neutral. In terms of [OH ], an acidic+ -7 -

solution has [OH ] < 1.0 × 10 M, a basic solution has [OH ] > 1.0 × 10 M, and a neutral solution- -7 - -7

has [OH ] = 1.0 × 10 M.- -7

a. [OH ] = = 1.0 × 10 M; The solution is neutral.- -7

b. [OH ] = = 1.5 × 10 M; The solution is acidic.- -11

c. [OH ] = = 5.3 × 10 M; The solution is basic.- -4

d. [OH ] = = 4.3 × 10 M; The solution is acidic.- -15

40. a. [H ] = = 2.8 × 10 M; basic+ -15

b. [H ] = = 1.0 × 10 M; acidic+ -6

c. [H ] = = 4.5 × 10 M; basic+ -12

d. [H ] = = 1.0 × 10 M; neutral+ -7

41. a. Since the value of the equilibrium constant increases as the temperature increases, the reaction isendothermic. In endothermic reactions, heat is a reactant so an increase in temperature (heat)shifts the reaction to produce more products and increases K in the process.

2 wb. H O(l) H (aq) + OH (aq) K = 5.47 × 10 = [H ][OH ] at 50°C+ - -14 + -

In pure water [H ] = [OH ], so 5.47 × 10 = [H ] , [H ] = 2.34 × 10 M = [OH ]+ - -14 + 2 + -7 -

2 w42. a. H O(l) H (aq) + OH (aq) K = 2.92 × 10 = [H ] [OH ]+ - -14 + -


In pure water [H ] = [OH ], so 2.92 × 10 = [H ] , [H ] = 1.71 × 10 M = [OH ]+ - -14 + 2 + -7 -

b. pH = -log [H ] = -log (1.71 × 10 ) = 6.767+ -7

wc. [H ] = K /[OH ] = 2.92 × 10 /0.10 = 2.9 × 10 M; pH = -log (2.9 × 10 ) =12.54+ - -14 -13 -13

43. pH = -log [H ]; pOH = -log [OH ]; At 25°C, pH + pOH = 14.00; For Exercise 13:39:+ -

a. pH = -log [H ] = -log (1.0 × 10 ) = 7.00; pOH = 14.00 - pH = 14.00 - 7.00 = 7.00+ -7

b. pH = -log (6.7 × 10 ) = 3.17; pOH = 14.00 - 3.17 = 10.83-4

c. pH = -log (1.9 × 10 ) = 10.72; pOH = 14.00 - 10.72 = 3.28-11

d. pH = -log (2.3) = -0.36; pOH = 14.00 - (-0.36) = 14.36

Note that pH is less than zero when [H ] is greater than 1.0 M (an extremely acidic solution).+

For Exercise 13.40:

a. pOH = -log [OH ] = -log (3.6) = -0.56; pH = 14.00 - pOH = 14.00 - (-0.56) = 14.56-

b. pOH = -log (9.7 × 10 ) = 8.01; pH = 14.00 - 8.01 = 5.99-9

c. pOH = -log (2.2 × 10 ) = 2.66; pH = 14.00 - 2.66 = 11.34-3

d. pOH = -log (1.0 × 10 ) = 7.00; pH = 14.00 - 7.00 = 7.00-7

Note that pH is greater than 14.00 when [OH ] is greater than 1.0 M (an extremely basic solution).-

44. a. [H ] = 10 , [H ] = 10 = 4.0 × 10 M+ -pH + -7.40 -8

pOH = 14.00 - pH = 14.00 - 7.40 = 6.60; [OH ] = 10 = 10 = 2.5 × 10 M- -pOH -6.60 -7

or [OH ] = = 2.5 × 10 M; This solution is basic since pH > 7.00.- -7

b. [H ] = 10 = 5 × 10 M; pOH = 14.00 - 15.3 = -1.3; [OH ] = 10 = 20 M; basic+ -15.3 -16 - -(-1.3)

c. [H ] = 10 = 10 M; pOH = 14.0 - (-1.0) = 15.0; [OH ] = 10 = 1 × 10 M; acidic+ -(-1.0) - -15.0 -15

d. [H ] = 10 = 6.3 × 10 M; pOH = 14.00 - 3.20 = 10.80; [OH ] = 10 = 1.6 × 10 M;+ -3.20 -4 - -10.80 -11

acidice. [OH ] = 10 = 1 × 10 M; pH = 14.0 - pOH = 14.0 - 5.0 = 9.0; [H ] = 10 = 1 × 10 M;- -5.0 -5 + -9.0 -9

basic

f. [OH ] = 10 = 2.5 × 10 M; pH = 14.00 - 9.60 = 4.40; [H ] = 10 = 4.0 × 10 M; acidic- -9.60 -10 + -4.40 -5

45. pOH = 14.00 - pH = 14.00 - 6.77 = 7.23; [H ] = 10 = 10 = 1.7 × 10 M+ -pH -6.77 -7


[OH ] = = 5.9 × 10 M or [OH ] = 10 = 10 = 5.9 × 10 M- -8 - -pOH -7.23 -8

The sample of milk is slightly acidic since the pH is less than 7.00 at 25°C.

46. pH = 14.00 - pOH = 14.00 - 5.74 = 8.26; [H ] = 10 = 10 = 5.5 × 10 M+ -pH -8.26 -9

[OH ] = = 1.8 × 10 M or [OH ] = 10 = 10 = 1.8 × 10 M- -6 - -pOH -5.74 -6

The solution of baking soda is basic since the pH is greater than 7.00 at 25°C.

Solutions of Acids

47. All the acids in this problem are strong acids that are always assumed to completely dissociate inwater. The general dissociation reaction for a strong acid is: HA(aq) ÷ H (aq) + A (aq) where A is+ - -

the conjugate base of the strong acid HA. For 0.250 M solutions of these strong acids, 0.250 M H+

and 0.250 M A are present when the acids completely dissociate. The amount of H donated from- +

2water will be insignificant in this problem since H O is a very weak acid.

4 2a. Major species present after dissociation = H , ClO and H O; + -

pH = -log [H ] = -log (0.250) = 0.602+

3 2b. Major species = H , NO and H O; pH = 0.602+ -

48. Strong acids are assumed to completely dissociate in water: HCl(aq) ÷ H (aq) + Cl (aq)+ -

a. A 0.10 M HCl solution gives 0.10 M H and 0.10 M Cl since HCl completely dissociates. The+ -

2amount of H from H O will be insignificant. pH = -log [H ] = -log (0.10) = 1.00+ +

2b. 5.0 M H is produced when 5.0 M HCl completely dissociates. The amount of H from H O will+ +

be insignificant. pH = -log (5.0) = -0.70 (Negative pH values just indicate very concentrated acidsolutions).

c. 1.0 × 10 M H is produced when 1.0 × 10 M HCl completely dissociates. If you take the-11 + -11

negative log of 1.0 × 10 this gives pH = 11.00. This is impossible! We dissolved an acid in-11

water and got a basic pH. What we must consider in this problem is that water by itself donates

21.0 × 10 M H . We can normally ignore the small amount of H from H O except when we have-7 + +

a very dilute solution of an acid (as is the case here). Therefore, the pH is that of neutral water(pH = 7.00) since the amount of HCl present is insignificant.

49. Both are strong acids.

0.0500 L × 0.050 mol/L = 2.5 × 10 mol HCl = 2.5 × 10 mol H + 2.5 × 10 mol Cl-3 -3 + -3 -

3 30.1500 L × 0.10 mol/L = 1.5 × 10 mol HNO = 1.5 × 10 mol H + 1.5 × 10 mol NO-2 -2 + -2 -

[H ] = = 0.088 M; [OH ] = = 1.1 × 10 M+ - -13


3[Cl ] = = 0.013 M; [NO ] = = 0.075 M - -

50. 90.0 × 10 L × = 0.450 mol H from HCl-3 +

330.0 × 10 L × = 0.240 mol H from HNO-3 +

[H ] = = 0.690 M; pH = -log (0.690) = 0.161+

pOH = 14.000 - 0.161 = 13.839; [OH ] = 10 = 1.45 × 10 M- -13.839 -14

51. [H ] = 10 = 10 = 3.2 × 10 M. Since HCl is a strong acid, a 3.2 × 10 M HCl solution will+ -pH -2.50 -3 -3

produce 3.2 × 10 M H giving a pH = 2.50.-3 +

3 352. [H ] = 10 = 10 = 7.9 × 10 M. Since HNO is a strong acid, a 7.9 × 10 M HNO solution is+ -pH -5.10 -6 -6

necessary to produce a pH = 5.10 solution.

2 a 2 a w 253. a. HNO (K = 4.0 × 10 ) and H O (K = K = 1.0 × 10 ) are the major species. HNO is a much-4 -14

2 2 astronger acid than H O so it is the major source of H . However, HNO is a weak acid (K < 1)+

so it only partially dissociates in water. We must solve an equilibrium problem to determine [H ].+

In the Solutions Guide, we will summarize the initial, change and equilibrium concentrations intoone table called the ICE table. Solving the weak acid problem:

2 2HNO H + NO+ -

Initial 0.250 M ~0 0

2x mol/L HNO dissociates to reach equilibrium

Change -x +x +xEquil. 0.250 -x x x

aK = = 4.0 × 10 = ; If we assume x << 0.250, then:-4

4.0 × 10 . , x = = 0.010 M -4

We must check the assumption:

All the assumptions are good. The H contribution from water (10 M) is negligible, and x is+ -7

small compared to 0.250 (percent error = 4.0%). If the percent error is less than 5% for anassumption, we will consider it a valid assumption (called the 5% rule). Finishing the problem:x = 0.010 M = [H ]; pH = -log(0.010) = 2.00+

3 2 a 2 a w 3 2b. CH CO H (K = 1.8 × 10 ) and H O (K = K = 1.0 × 10 ) are the major species. CH CO H-5 -14

is the major source of H . Solving the weak acid problem:+


3 2 3 2CH CO H H + CH CO + -


3 2x mol/L CH CO H dissociates to reach equilibriumChange -x +x +xEquil. 0.250 - x x x

aK = = 1.8 × 10 = (assuming x << 0.250)-5

x = 2.1 × 10 M; Checking assumption: × 100= 0.84%. Assumptions good.-3

[H ] = x = 2.1 × 10 M; pH = -log (2.1 × 10 ) = 2.68+ -3 -3

6 5 a 2 a w54. a. HOC H (K = 1.6 × 10 ) and H O (K = K = 1.0 × 10 ) are the major species. The major-10 -14

6 5equilibrium is the dissociation of HOC H . Solving the weak acid problem:

6 5 6 5HOC H H + OC H+ -


6 5x mol/L HOC H dissociates to reach equilibriumChange -x +x +xEquil. 0.250 - x x x

aK = 1.6 × 10 = (assuming x << 0.250)-10

x = [H ] = 6.3 × 10 M; Checking assumption: x is 2.5 × 10 % of 0.250, so assumption is valid+ -6 -3

by the 5% rule.

pH = -log(6.3 × 10 ) = 5.20-6

a 2b. HCN (K = 6.2 × 10 ) and H O are the major species. HCN is the major source of H .-10 +

HCN H + CN+ -

Initial 0.250 M ~0 0x mol/L HCN dissociates to reach equilibrium

Change -x +x +xEquil. 0.250 - x x x

aK = 6.2 × 10 = (assuming x << 0.250)-10


x = [H ] = 1.2 × 10 M; Checking assumption: x is 4.8 × 10 % of 0.250+ -5 -3

Assumptions good. pH = -log (1.2 × 10 ) = 4.92-5

2 3 255. 20.0 mL glacial acetic acid × = 0.350 mol HC H O

2 3 2Initial concentration of HC H O = = 1.40 M

2 3 2 2 3 2 aHC H O H + C H O K = 1.8 × 10+ - -5

Initial 1.40 M ~0 0

2 3 2x mol/L HC H O dissociates to reach equilibriumChange -x +x +xEquil. 1.40 - x x x

aK = 1.8 × 10 = -5

x = [H ] = 5.0 × 10 M; pH = 2.30 Assumptions good (x is 0.36% of 1.40).+ -3

3 5 2 a 2 a w 3 5 256. HC H O (K = 1.3 × 10 ) and H O (K = K = 1.0 × 10 ) are the major species present. HC H O-5 -14

3 5 2 2will be the dominant producer of H since HC H O is a stronger acid than H O. Solving the weak+

acid problem:

3 5 2 3 5 2HC H O H + C H O+ -



aK = 1.3 × 10 = -5

x = [H ] = 1.1 × 10 M; pH = -log (1.1 × 10 ) = 2.96+ -3 -3

Assumption follows the 5% rule (x is 1.1% of 0.100).

3 5 2 w[H ] = [C H O ] = 1.1 × 10 M; [OH ] = K /[H ] = 9.1 × 10 M+ - -3 - + -12

3 5 2[HC H O ] = 0.100 - 1.1 × 10 = 0.099 M-3

Percent dissociation = × 100 = 1.1%


57. This is a weak acid in water. Solving the weak acid problem:

a HF H + F K = 7.2 × 10+ - -4

Initial 0.020 M ~0 0x mol/L HF dissociates to reach equilibrium


aK = 7.2 × 10 = (assuming x << 0.020)-4

x = [H ] = 3.8 × 10 M; Check assumptions: + -3

The assumption x << 0.020 is not good (x is more than 5% of 0.020). We must solve x /(0.020 - x) = 7.2 × 10 exactly by using either the quadratic formula or by the method of2 -4

successive approximations (see Appendix 1.4 of text). Using successive approximations, we let0.016 M be a new approximation for [HF]. That is, in the denominator, try x = 0.0038 (the valueof x we calculated making the normal assumption), so 0.020 - 0.0038 = 0.016, then solve for anew value of x in the numerator.

= 7.2 × 10 , x = 3.4 × 10-4 -3

We use this new value of x to further refine our estimate of [HF], i.e., 0.020 - x = 0.020 - 0.0034 = 0.0166 (carry extra significant figure).

= 7.2 × 10 , x = 3.5 × 10-4 -3

We repeat until we get an answer that repeats itself. This would be the same answer we wouldget solving exactly using the quadratic equation. In this case it is: x = 3.5 × 10-3

wSo: [H ] = [F ] = x = 3.5 × 10 M; [OH ] = K /[H ] = 2.9 × 10 M+ - -3 - + -12

[HF] = 0.020 - x = 0.020 - 0.0035 = 0.017 M; pH = 2.46

Note: When the 5% assumption fails, use whichever method you are most comfortable with tosolve exactly. The method of successive approximations is probably fastest when the percent erroris less than ~25% (unless you have a calculator that can solve quadratic equations).

3 2 3 a58. Major species: HIO , H O; Major source of H : HIO (a weak acid, K = 0.17)+

3 3HIO H + IO+ -

Initial 0.20 M ~0 0

3x mol/L HIO dissociates to reach equilibriumChange -x +x +xEquil. 0.20 - x x x


aK = 0.17 = , x = 0.18; Check assumption.

Assumption is horrible (x is 90% of 0.20). When the assumption is this poor, it is generally quickestto solve exactly using the quadratic formula (see Appendix 1.4 in text). The method of successiveapproximations will require many trials to finally converge on the answer. For this problem, 5 trialswere required. Using the quadratic formula and carrying extra significant figures:

0.17 = , x = 0.17(0.20 - x), x + 0.17 x - 0.034 = 02 2

x = , x = 0.12 or -0.29

Only x = 0.12 makes sense. x = 0.12 M = [H ]; pH = -log (0.12) = 0.92+

2 2 2 a 2 2 2 259. Major species: HC H ClO (K = 1.35 × 10 ) and H O; Major source of H : HC H ClO-3 +

2 2 2 2 2 2HC H ClO H + C H ClO+ -

Initial 0.10 M ~0 0

2 2 2x mol/L HC H ClO dissociates to reach equilibriumChange -x +x +xEquil. 0.10 - x x x

aK = 1.35 × 10 = , x = 1.2 × 10 M-3 -2

Checking the assumptions finds that x is 12% of 0.10 which fails the 5% rule. We must solve1.35 × 10 = x /(0.10 - x) exactly using either the method of successive approximations or the-3 2

quadratic equation. Using either method gives x = [H ] = 1.1 × 10 M. pH = -log [H ] = -log (1.1+ -2 +

× 10 ) = 1.96.-2

2 2 a60. HClO H + ClO K = 1.2 × 10 + - -2

Initial 0.22 M ~0 0

2x mol/L HClO dissociates to reach equilibriumChange -x +x +xEquil. 0.22 - x x x

aK = 1.2 × 10 = , x = 5.1 × 10-2 -2

The assumption that x is small is not good (x is 23% of 0.22). Using the method of successiveapproximations and carrying extra significant figures:


= 1.2 × 10 , x = 4.5 × 10-2 -2

= 1.2 × 10 , x = 4.6 × 10 ; x = 4.6 × 10 repeats-2 -2 -2

2[H ] = [ClO ] = x = 4.6 × 10 M; % dissociation = × 100 = 21%+ - -2

61. a. HCl is a strong acid. It will produce 0.10 M H . HOCl is a weak acid. Let's consider the+

equilibrium:

aHOCl H + OCl K = 3.5 × 10+ - -8

Initial 0.10 M 0.10 M 0x mol/L HOCl dissociates to reach equilibrium

Change -x +x +xEquil. 0.10 - x 0.10 + x x

aK = 3.5 × 10 = , x = 3.5 × 10 M-8 -8

Assumptions are great (x is 3.5 × 10 % of 0.10). We are really assuming that HCl is the only-5

important source of H , which it is. The [H ] contribution from HOCl, x, is negligible. Therefore,+ +

[H ] = 0.10 M; pH = 1.00+

3b. HNO is a strong acid, giving an initial concentration of H equal to 0.050 M. Consider the+

equilibrium:

2 3 2 2 3 2 aHC H O H + C H O K = 1.8 × 10+ - -5

Initial 0.50 M 0.050 M 0

2 3 2x mol/L HC H O dissociates to reach equilibriumChange -x +x +xEquil. 0.50 - x 0.050 + x x

aK = 1.8 × 10 =-5

x = 1.8 × 10 ; Assumptions are good (well within the 5% rule).-4

[H ] = 0.050 + x = 0.050 M and pH = 1.30+

6 5 a62. HF and HOC H are both weak acids with K values of 7.2 × 10 and 1.6 × 10 , respectively. Since-4 -10

a a 6 5the K value for HF is much greater than the K value for HOC H , HF will be the dominant producer

6 5of H (we can ignore the amount of H produced from HOC H since it will be insignificant).+ +

HF H + F+ -




aK = 7.2 × 10 = -4

x = [H ] = 2.7 × 10 M; pH = -log (2.7 × 10 ) = 1.57 Assumptions good.+ -2 -2

6 5 6 5 6 5Solving for [OC H ] using HOC H º H + OC H equilibrium: - + -

a 6 5K = 1.6 × 10 = , [OC H ] = 5.9 × 10 M-10 - -9

6 5Note that this answer indicates that only 5.9 × 10 M HOC H dissociates, which indicates that-9

HF is truly the only significant producer of H in this solution.+

2 3 263. In all parts of this problem, acetic acid (HC H O ) is the best weak acid present. We must solve aweak acid problem.

2 3 2 2 3 2a. HC H O H + C H O+ -

Initial 0.50 M ~0 0


aK = 1.8 × 10 = -5

2 3 2x = [H ] = [C H O ] = 3.0 × 10 M Assumptions good.+ - -3

Percent dissociation = × 100 = × 100 = 0.60%

b. The setups for solutions b and c are similar to solution a except the final equation is slightly

2 3 2different, reflecting the new concentration of HC H O .

aK = 1.8 × 10 = -5

2 3 2x = [H ] = [C H O ] = 9.5 × 10 M Assumptions good.+ - -4

% dissociation = × 100 = 1.9%

a c. K = 1.8 × 10 = -5

2 3 2x = [H ] = [C H O ] = 3.0 × 10 M; Check assumptions.+ - -4

Assumption that x is negligible is borderline (6.0% error). We should solve exactly. Using the


method of successive approximations (see Appendix 1.4 of text):

1.8 × 10 = , x = 2.9 × 10-5 -4

Next trial also gives x = 2.9 × 10 .-4


d. As we dilute a solution, all concentrations decrease. Dilution will shift the equilibrium to the sidewith the greater number of particles. For example, suppose we double the volume of anequilibrium mixture of a weak acid by adding water, then:

Q =

aQ < K , so the equilibrium shifts to the right or towards a greater percent dissociation.

e. [H ] depends on the initial concentration of weak acid and on how much weak acid dissociates.+

For solutions a-c the initial concentration of acid decreases more rapidly than the percentdissociation increases. Thus, [H ] decreases.+

364. a. HNO is a strong acid; it is assumed 100% dissociated in solution.

2 2 ab. HNO H + NO K = 4.0 × 10+ - -4

Initial 0.20 M ~0 0

2x mol/L HNO dissociates to reach equilibriumChange -x +x +xEquil. 0.20 - x x x

aK = 4.0 × 10 = -4

2 x = [H ] = [NO ] = 8.9 × 10 M; Assumptions good. + - -3


6 5 6 5 ac. HOC H H + OC H K = 1.6 × 10+ - -10


Initial 0.20 M ~0 0

6 5x mol/L HOC H dissociates to reach equilibriumChange -x +x +xEquil. 0.20 - x x x

aK = 1.6 × 10 = -10

6 5 x = [H ] = [OC H ] = 5.7 × 10 M; Assumptions good. + - -6

% dissociation = × 100 = 2.9 × 10 %-3

d. For the same initial concentration, the percent dissociation increases as the strength of the acid

aincreases (as K increases).

65. Let HX symbolize the weak acid. Setup the problem like a typical weak acid equilibrium problem.

HX H + X+ -

Initial 0.15 M ~0 0x mol/L HX dissociates to reach equilibrium


If the acid is 3.0% dissociated, then x = [H ] is 3.0% of 0.15: x = 0.030 × (0.15 M) = +

a4.5 × 10 M. Now that we know the value of x, we can solve for K .-3

aK = = 1.4 × 10-4

66. HF H + F+ -



aK = ; x = [H ] = [F ] = 0.081 × (0.100 M) = 8.1 × 10 M+ - -3

a[HF] = 0.100 - 8.1 × 10 = 0.092 M; K = = 7.1 × 10-3 -4

a67. Setup the problem using the K equilibrium reaction for HOCN.

HOCN H + OCN+ -

Initial 0.0100 M ~0 0x mol/L HOBr dissociates to reach equilibrium



aK = ; Since pH = 2.77, then: x = [H ] = 10 = 10 = 1.7 × 10 M+ -pH -2.77 -3

aK = = 3.5 × 10-4

468. HClO is a strong acid with [H ] = 0.040 M. This equals the [H ] in the trichloroacetic acid solution.+ +

a 3 2Now setup the problem using the K equilibrium reaction for CCl CO H.

3 2 3 2CCl CO H H + CCl CO+ -

Initial 0.050 M ~0 0Equil. 0.050 - x x x

aK = ; x = [H ] = 4.0 × 10 M+ -2

aK = = 0.16

269. Major species: HCOOH and H O; Major source of H : HCOOH+

HCOOH H + HCOO+ -

oInitial C ~0 0 where C = [HCOOH]x mol/L HCOOH dissociates to reach equilibrium

Change -x +x +xEquil. C - x x x

aK = 1.8 × 10 = where x = [H ]-4 +

1.8 × 10 = ; Since pH = 2.70, then: [H ] = 10 = 2.0 × 10 M-4 + -2.70 -3

1.8 × 10 = , C - (2.0 × 10 ) = , C = 2.4 × 10 M-4 -3 -2

A 0.024 M formic acid solution will have pH = 2.70.

o a70. [HA] = = 0.50 mol/L; Solve using the K equilibrium reaction.

HA H + A+ -



aK = ; In this problem, [HA] = 0.45 M so:

a[HA] = 0.45 M = 0.50 M - x, x = 0.05 M; K = = 6 × 10-3

Solutions of Bases

3 2 4 b71. a. NH (aq) + H O(l) º NH (aq) + OH (aq) K = + -

5 5 2 5 5 bb. C H N(aq) + H O(l) º C H NH (aq) + OH (aq) K = + -

6 5 2 2 6 5 3 b72. a. C H NH (aq) + H O(l) º C H NH (aq) + OH (aq) K = + -

3 2 2 3 2 2 bb. (CH ) NH(aq) + H O(l) º (CH ) NH (aq) + OH (aq) K = + -

3 b w 373. NO : K << K since HNO is a strong acid. All conjugate bases of strong acids have no base-

2 b w 3 b 5 5 bstrength. H O: K = K = 1.0 × 10 ; NH : K = 1.8 × 10 ; C H N: K = 1.7 × 10-14 -5 -9

3 5 5 2 3 bNH > C H N > H O > NO (As K increases, base strength increases.)-

74. Excluding water, these are the conjugate acids of the bases in the previous exercise. In general, the

4 5 5stronger the base, the weaker the conjugate acid. Note: Even though NH and C H NH are+ +

a wconjugate acids of weak bases, they are still weak acids with K values between K and 1. Prove this

a 4 5 5 a w bto yourself by calculating the K values for NH and C H NH (K = K /K ).+ +

3 5 5 4 2HNO > C H NH > NH > H O+ +

3 3 3 275. a. NH b. NH c. OH d. CH NH-

bThe base with the largest K value is the strongest base. OH is the strongest base possible in water.-

3 4 476. a. HNO b. NH c. NH+ +

a a 4The acid with the largest K value is the strongest acid. To calculate K values for NH and+

3 3 a w b b 3 3 2CH NH , use K = K /K where K refers to the bases NH or CH NH .+

77. NaOH(aq) ÷ Na (aq) + OH (aq); NaOH is a strong base which completely dissociates into Na and+ - +

OH . The initial concentration of NaOH will equal the concentration of OH donated by NaOH.- -

a. [OH ] = 0.10 M; pOH = -log[OH ] = -log(0.10) = 1.00- -

pH = 14.00 - pOH = 14.00 - 1.00 = 13.00

2 2Note that H O is also present, but the amount of OH produced by H O will be insignificant-

compared to the 0.10 M OH produced from the NaOH.-

b. The [OH ] concentration donated by the NaOH is 1.0 × 10 M. Water by itself donates - -10


1.0 × 10 M. In this problem, water is the major OH contributor and [OH ] = 1.0 × 10 M.-7 - - -7

pOH = -log (1.0 × 10 ) = 7.00; pH = 14.00 - 7.00 = 7.00-7

c. [OH ] = 2.0 M; pOH = -log (2.0) = -0.30; pH = 14.00 - (-0.30) = 14.30-

2 278. a. Ca(OH) ÷ Ca + 2 OH ; Ca(OH) is a strong base and dissociates completely.2+ -

[OH ] = 2(0.00040) = 8.0 × 10 M; pOH = -log [OH ] = 3.10; pH = 14.00 - pOH = 10.90- -4 -

b. = 0.45 mol KOH/L

KOH is a strong base, so [OH ] = 0.45 M; pOH = -log (0.45) = 0.35; pH = 13.65-

c. = 3.750 M; NaOH is a strong base, so [OH ] = 3.750 M.-

pOH = -log (3.750) = -0.5740 and pH = 14.0000 - (-0.5740) = 14.5740

Although we are justified in calculating the answer to four decimal places, in reality pH values aregenerally measured to two decimal places, sometimes three.

279. a. Major species: K , OH , H O (KOH is a strong base.)+ -

[OH ] = 0.015 M, pOH = -log (0.015) = 1.82; pH = 14.00 - pOH = 12.18-

2 2b. Major species: Ba , OH , H O; Ba(OH) (aq) ÷ Ba (aq) + 2 OH (aq); Since each mol of the2+ - 2+ -

2strong base Ba(OH) dissolves in water to produce two mol OH , then [OH ] = 2(0.015 M) - -

= 0.030 M.

pOH = -log (0.030) = 1.52; pH =14.00 - 1.52 = 12.48280. a. Major species: Na , Li , OH , H O (NaOH and LiOH are both strong bases.)+ + -

[OH ] = 0.050 + 0.050 = 0.100 M; pOH = 1.000; pH = 13.000-

2 2 2b. Major species: Ba , Rb , OH , H O; Both Ba(OH) and RbOH are strong bases and Ba(OH)2+ + -

2donates 2 mol OH per mol Ba(OH) .-

[OH ] = 2(0.0010) + 0.020 = 0.022 M; pOH = -log (0.022) = 1.66; pH = 12.34-

81. pH = 10.50; pOH = 14.00 - 10.50 = 3.50; [OH ] = 10 = 3.2 × 10 M- -3.50 -4

KOH(aq) º K (aq) + OH (aq); Since KOH is a strong base, a 3.2 × 10 M KOH solution will+ - -4

produce a pH = 10.50 solution.

82. pH = 10.50; pOH = 14.00 - 10.50 = 3.50; [OH ] = 10 = 3.2 × 10 M - -3.50 -4

2 2 2Sr(OH) (aq) ÷ Sr (aq) + 2 OH (aq); Sr(OH) donates two mol OH per mol Sr(OH) .2+ - -

2 2[Sr(OH) ] = 3.2 × 10 M OH × = 1.6 × 10 M Sr(OH)-4 - -4


2A 1.6 × 10 M Sr(OH) solution will produce a pH = 10.50 solution.-4

3 b 3 2 b w83. NH is a weak base with K = 1.8 × 10 . The major species present will be NH and H O (K = K-5

3 b 2 3= 1.0 × 10 ). Since NH has a much larger K value compared to H O, NH is the stronger base-14

3present and will be the major producer of OH . To determine the amount of OH produced from NH ,- -

we must perform an equilibrium calculation.

3 2 4NH (aq) + H O(l) NH (aq) + OH (aq)+ -

Initial 0.150 M 0 ~0

3 2x mol/L NH reacts with H O to reach equilibriumChange -x +x +xEquil. 0.150 - x x x

bK = 1.8 × 10 = (assuming x << 0.150)-5

x = [OH ] = 1.6 × 10 M; Check assumptions: x is 1.1% of 0.150 so the assumption 0.150 - x .- -3

0.150 is valid by the 5% rule. Also, the contribution of OH from water will be insignificant (which-

will usually be the case). Finishing the problem, pOH = -log [OH ] = -log (1.6 × 10 M) = 2.80; pH- -3

= 14.00 - pOH = 14.00 - 2.80 = 11.20.

2 2 b 2 b w 2 284. Major species: H NNH (K = 3.0 × 10 ) and H O (K = K = 1.0 × 10 ); The weak base H NNH-6 -14

will dominate OH production. We must perform a weak base equilibrium calculation.-

2 2 2 2 3 bH NNH + H O H NNH + OH K = 3.0 × 10+ - -6

Initial 2.0 M 0 ~0

2 2 2x mol/L H NNH reacts with H O to reach equilibriumChange -x +x +xEquil. 2.0 - x x x

bK = 3.0 × 10 = (assuming x << 2.0)-6

x = [OH ] = 2.4 × 10 M; pOH = 2.62; pH = 11.38 Assumptions good (x is 0.12% of 2.0).- -3

2 3 2 2[H NNH ] = 2.4 × 10 M; [H NNH ] = 2.0 M; [H ] = 10 = 4.2 × 10 M+ -3 + -11..38 -12

85. These are solutions of weak bases in water. We must solve the equilibrium weak base problem.

2 5 3 2 2 5 3 ba. (C H ) N + H O (C H ) NH + OH K = 4.0 × 10+ - -4

Initial 0.20 M 0 ~0

2 5 3 2x mol/L of (C H ) N reacts with H O to reach equilibriumChange -x +x +x


Equil. 0.20 - x x x

bK = 4.0 × 10 = , x = [OH ] = 8.9 × 10 M-4 - -3

Assumptions good (x is 4.5% of 0.20). [OH ] = 8.9 × 10 M- -3

[H ] = = 1.1 × 10 M; pH = 11.96+ -12

2 2 3 bb. HONH + H O HONH + OH K = 1.1 × 10+ - -8

Initial 0.20 M 0 ~0Equil. 0.20 - x x x

bK = 1.1 × 10 = , x = [OH ] = 4.7 × 10 M; Assumptions good.-8 - -5

[H ] = 2.1 × 10 M; pH = 9.68+ -10

86. These are solutions of weak bases in water.

6 5 2 2 6 5 3 ba. C H NH + H O C H NH + OH K = 3.8 × 10+ - -10

Initial 0.20 M 0 ~0

6 5 2 2x mol/L of C H NH reacts with H O to reach equilibriumChange -x +x +xEquil. 0.20 - x x x

3.8 × 10 = , x = [OH ] = 8.7 × 10 M; Assumptions good.-10 - -6

w[H ] = K /[OH ] = 1.1 × 10 M; pH = 8.96+ - -9

5 5 2 5 5 bb. C H N + H O C H NH + OH K = 1.7 × 10+ - -9


bK = 1.7 × 10 = , x = 1.8 × 10 M; Assumptions good.-9 -5


[OH ] = 1.8 × 10 M; [H ] = 5.6 × 10 M; pH = 9.25- -5 + -10

87. This is a solution of a weak base in water. We must solve the weak base equilibrium problem.

2 5 2 2 2 5 3 bC H NH + H O C H NH + OH K = 5.6 × 10+ - -4

Initial 0.20 M 0 ~0

2 5 2 2x mol/L C H NH reacts with H O to reach equilibriumChange -x +x +xEquil. 0.20 - x x x

bK = (assuming x << 0.20)

x = 1.1 × 10 ; Checking assumption: × 100 = 5.5%-2

Assumption fails the 5% rule. We must solve exactly using either the quadratic equation or themethod of successive approximations (see Appendix 1.4 of the text). Using successiveapproximations and carrying extra significant figures:

= 5.6 × 10 , x = 1.0 × 10 M (consistent answer)-4 -2

x = [OH ] = 1.0 × 10 M; [H ] = = 1.0 × 10 M; pH = 12.00- -2 + -12

2 5 2 2 2 5 2 2 b88. (C H ) NH + H O (C H ) NH + OH K = 1.3 × 10+ - -3


2 5 2 2x mol/L (C H ) NH reacts with H O to reach equilibriumChange -x +x +xEquil. 0.050 - x x x

bK = 1.3 × 10 = (assuming x << 0.050)-3

x = 8.1 × 10 ; Assumption is bad (x is 16% of 0.20).-3

Using successive approximations:

1.3 × 10 = , x = 7.4 × 10-3 -3

1.3 × 10 = , x = 7.4 × 10 (consistent answer)-3 -3

w[OH ] = x = 7.4 × 10 M; [H ] = K /[OH ] = 1.4 × 10 M; pH = 11.85- -3 + - -12


89. To solve for percent ionization, just solve the weak base equilibrium problem.

3 2 4 ba. NH + H O NH + OH K = 1.8 × 10+ - -5


bK = 1.8 × 10 = , x = [OH ] = 1.3 × 10 M; Assumptions good.-5 - -3

Percent ionization = × 100 = 1.3%

3 2 4b. NH + H O NH + OH+ -




Note: For the same base, the percent ionization increases as the initial concentration of base decreases.

2 2 3 b90. a. HONH + H O HONH + OH K = 1.1 × 10+ - -8




3 2 2 3 3 bb. CH NH + H O CH NH + OH K = 4.38 × 10+ - -4


4.38 × 10 = , x = 6.6 × 10 ; Assumption fails the 5% rule (x is 6.6% -4 -3

of 0.10).

Using successive approximations and carrying extra significant figures:

= 4.38 × 10 , x = 6.4 × 10 (consistent answer)-4 -3



18 21 3 b91. Let cod = codeine, C H NO ; using the K reaction to solve:

2cod + H O codH + OH+ -

Initial 1.7 × 10 M 0 ~0-3

2x mol/L codeine reacts with H O to reach equilibriumChange -x +x +xEquil. 1.7 × 10 - x x x-3

bK = ; Since pH = 9.59, then pOH = 14.00 - 9.59 = 4.41.

b[OH ] = x = 10 = 3.9 × 10 M; K = = 9.2 × 10- -4.41 -5 -7

b 4 892. Using the K reaction to solve where PY = pyrrolidine, C H NH:

2 PY + H O PYH + OH+ -

Initial 1.00 × 10 M 0 ~0-3

Equil. 1.00 × 10 - x x x-3

bK = ; Since pH = 10.82: pOH = 3.18 and [OH ] = x = 10 = 6.6 × 10 M- -3.18 -4

b K = = 1.3 × 10-3

Polyprotic Acids

2 3 393. H SO (aq) º HSO (aq) + H (aq) = - +

3 3HSO (aq) º SO (aq) + H (aq) = - 2- +

3 6 5 7 2 6 5 794. H C H O (aq) º H C H O (aq) + H (aq) = - +

2 6 5 7 6 5 7H C H O (aq) º HC H O (aq) + H (aq) = - 2- +

6 5 7 6 5 7HC H O (aq) º C H O (aq) + H (aq) = 2- 3- +


95. In both these polyprotic acid problems, the dominate equilibrium is the reaction. The amount of

aH produced from the subsequent K reactions will be minimal since they are all have much smaller+

aK values.

3 4 2 4a. H PO H + H PO = 7.5 × 10+ - -3

Initial 0.10 M ~0 0

3 4x mol/L H PO dissociates to reach equilibriumChange -x +x +xEquil. 0.10 - x x x

= 7.5 × 10 = , x = 2.7 × 10-3 -2

Assumption is bad (x is 27% of 0.10). Using successive approximations:

= 7.5 × 10 , x = 2.3 × 10 ; = 7.5 × 10 , x = 2.4 × 10-3 -2 -3 -2

(consistent answer)

x = [H ] = 2.4 × 10 M; pH = -log (2.4 × 10 ) = 1.62+ -2 -2

2 3 3b. H CO H + HCO = 4.3 × 10+ - -7


= 4.3 × 10 = -7

x = [H ] = 2.1 × 10 M; pH = 3.68; Assumptions good.+ -4

96. The reactions are:

3 4 2 4H AsO º H + H AsO = 5 × 10+ - -3

2 4 4H AsO º H + HAsO = 8 × 10- + 2- -8

4 4HAsO º H + AsO = 6 × 102- + 3- -10

aWe will deal with the reactions in order of importance, beginning with the largest K , .

3 4 2 4H AsO H + H AsO = 5 × 10 = + - -3


5 × 10 = , x = 2.9 × 10 = 3 × 10 M (By using successive approximations-3 -2 -2


or the quadratic formula.)

2 4 3 4[H ] = [H AsO ] = 3 × 10 M; [H AsO ] = 0.20 - 0.03 = 0.17 M+ - -2

Since = 8 × 10 is much smaller than the value, very little of -8

2 4 4 3 4 2 4H AsO (and HAsO ) dissociates compared to H AsO . Therefore, [H ] and [H AsO ] will not- 2- + -

change significantly by the reaction. Using the previously calculated concentrations of H and+

2 4 4H AsO to calculate the concentration of HAsO :- 2-

48 × 10 = , [HAsO ] = 8 × 10 M-8 2- -8

a2Assumption that the K reaction does not change [H ] and [ ] is good. We repeat the process+

4using to get [AsO ].3-

= 6 × 10 = -10

4[AsO ] = 1.6 × 10 . 2 × 10 M Assumption good.3- -15 -15

3 4So in 0.20 M analytical concentration of H AsO :

3 4 2 4[H AsO ] = 0.17 M; [H ] = [H AsO ] = 3 × 10 M+ - -2

4 4[HAsO ] = 8 × 10 M; [AsO ] = 2 × 10 M2- -8 3- -15

w[OH ] = K /[H ] = 3 × 10 M- + -13

2 4 2 4 497. The dominant H producer is the strong acid H SO . A 2.0 M H SO solution produces 2.0 M HSO+ -

4and 2.0 M H . However, HSO is a weak acid which could also add H to the solution.+ - +

4 4HSO H + SO - + 2-

Initial 2.0 M 2.0 M 0

4x mol/L HSO dissociates to reach equilibrium-


= 1.2 × 10 = , x = 1.2 × 10-2 -2

Since x is 0.60% of 2.0, the assumption is valid by the 5% rule. The amount of additional H from+


4HSO is 1.2 × 10 . The total amount of H present is:- -2 +

[H ] = 2.0 + 1.2 × 10 = 2.0 M; pH = -log (2.0) = -0.30+ -2

4Note: In this problem, H from HSO could have been ignored. However, this is not usually the case,+ -

2 4especially in more dilute solutions of H SO .

2 4 498. For H SO , the first dissociation occurs to completion. The hydrogen sulfate ion, HSO , is a weak-

acid with = 1.2 × 10 . We will consider this equilibrium for additional H production:-2 +

4 4HSO H + SO- + 2-

Initial 0.0050 M 0.0050 M 0

4x mol/L HSO dissociates to reach equilibrium-


= 0.012 = . x, x = 0.012; Assumption is horrible (240% error).

Using the quadratic formula:

6.0 × 10 - 0.012 x = x + 0.0050 x, x + 0.017 x - 6.0 × 10 = 0-5 2 2 -5

x = , x = 3.0 × 10 M-3

[H ] = 0.0050 + x = 0.0050 + 0.0030 = 0.0080 M; pH = 2.10+

2 4 4Note: We had to consider both H SO and HSO for H production in this problem.- +

Acid-Base Properties of Salts

99. One difficult aspect of acid-base chemistry is recognizing what types of species are present in solution,i.e., whether a species is a strong acid, strong base, weak acid, weak base or a neutral species. Beloware some ideas and generalizations to keep in mind that will help in recognizing types of speciespresent.

3 4 2 4a. Memorize the following strong acids: HCl, HBr, HI, HNO , HClO and H SO

2 2b. Memorize the following strong bases: LiOH, NaOH, KOH, RbOH, Ca(OH) , Sr(OH) and

2Ba(OH)

a wc. All weak acids have a K value less than 1 but greater than K . Some weak acids are in Table

b w14.2 of the text. All weak bases have a K value less than 1 but greater than K . Some weakbases are in Table 14.3 of the text.

bd. All conjugate bases of weak acids are weak bases, i.e., all have a K value less than 1 but greater

wthan K . Some examples of these are the conjugate bases of the weak acids in Table 14.2 of thetext.

ae. All conjugate acids of weak bases are weak acids, i.e., all have a K value less than 1 but greater

wthan K . Some examples of these are the conjugate acids of the weak bases in Table 14.3 of the


text.f. Alkali metal ions (Li , Na , K , Rb , Cs ) and heavier alkaline earth metal ions (Ca , Sr , Ba )+ + + + + 2+ 2+ 2+

have no acidic or basic properties in water.

3 4 4g. All conjugate bases of strong acids (Cl , Br , I , NO , ClO , HSO ) have no basic properties in- - - - - -

b w 4water (K << K ) and only HSO has any acidic properties in water.-

Lets apply these ideas to this problem to see what type of species are present. The letters inparenthesis is/are the generalization(s) above which identifies the species.

KOH: strong base (b)KCl: neutral; K and Cl have no acidic/basic properties (f and g).+ -

bKCN: CN is a weak base, K = 1.0 × 10 /6.2 × 10 = 1.6 × 10 (c and d). Ignore K (f).- -14 -10 -5 +

4 4 aNH Cl: NH is a weak acid, K = 5.6 × 10 (c and e). Ignore Cl (g).+ -10 -

HCl: strong acid (a)

The most acidic solution will be the strong acid followed by the weak acid. The most basic solutionwill be the strong base followed by the weak base. The KCl solution will be between the acidic andbasic solutions at pH = 7.00.

4Most acidic ÷ most basic; HCl > NH Cl > KCl > KCN > KOH100. See Exercise 14.99 for some generalizations on acid-base properties of salts. The letters in parenthesis

is/are the generalization(s) listed in Exercise 14.99 which identifies the species.

2CaBr : neutral; Ca and Br have no acidic/basic properties (f and g).2+ -

2 2 bKNO : NO is a weak base, K = 1.0 × 10 /4.0 × 10 = 2.5 × 10 (c and d). - -14 -4 -11

Ignore K (f).+

4HClO : strong acid (a)

2 aHNO : weak acid, K = 4.0 × 10 (c)-4

3 4 3 aHONH ClO : HONH is a weak acid, K = 1.0 × 10 /1.1 × 10 = 9.1 × 10 (c and e). + -14 -8 -7

4Ignore ClO (g).-

a bUsing the information above (identity and K or K values), the ordering is:

4 2 3 4 2 2most acidic ÷ most basic: HClO > HNO > HONH ClO > CaBr > KNO

a101. From the K values, acetic acid is a stronger acid than hypochlorous acid. Conversely, the conjugate

2 3 2base of acetic acid, C H O , will be a weaker base than the conjugate base of hypochlorous acid, OCl .- -

2 3 2Thus, the hypochlorite ion, OCl , is a stronger base than the acetate ion, C H O . In general, the- -

w astronger the acid, the weaker the conjugate base. This statement comes from the relationship K = K

b× K , which holds for all conjugate acid-base pairs.

3 b 3 2 3102. Since NH is a weaker base (smaller K value) than CH NH , the conjugate acid of NH will be a

3 2 4 3 3stronger acid than the conjugate acid of CH NH . Thus, NH is a stronger acid than CH NH .+ +

3 3 3103. NaN ÷ Na + N ; Azide, N , is a weak base since it is the conjugate base of a weak acid. All+ - -

w bconjugate bases of weak acids are weak bases (K < K < 1). Ignore Na .+

3 2 3 b N + H O HN + OH K = = 5.3 × 10- - -10



3 2x mol/L of N reacts with H O to reach equilibrium-


bK = = 5.3 × 10 = (assuming x << 0.010)-10

x = [OH ] = 2.3 × 10 M; [H ] = = 4.3 × 10 M Assumptions good.- -6 + -9

3 3[HN ] = [OH ] = 2.3 × 10 M; [Na ] = 0.010 M; [N ] = 0.010 - 2.3 × 10 = 0.010 M - -6 + - -6

2 5 3 2 5 3 2 5 3 2 5 2104. C H NH Cl ÷ C H NH + Cl ; C H NH is the conjugate acid of the weak base C H NH + - +

b 2 5 3(K = 5.6 × 10 ). As is true for all conjugate acids of weak bases, C H NH is a weak acid. Cl has-4 + -

no basic (or acidic) properties. Ignore Cl . Solving the weak acid problem:-

2 5 3 2 5 2 a wC H NH C H NH + H K = K /5.6 × 10 = 1.8 × 10+ + -4 -11

Initial 0.25 M 0 ~0

2 5 3x mol/L C H NH dissociates to reach equilibrium+


aK = 1.8 × 10 = (assuming x << 0.25)-11


2 5 2 2 5 3[C H NH ] = [H ] = 2.1 × 10 M; [C H NH ] = 0.25 M; [Cl ] = 0.25 M+ -6 + -

w[OH ] = K /[H ] = 4.8 × 10 M- + -9

3 3 3 3 3 3105. a. CH NH Cl ÷ CH NH + Cl : CH NH is a weak acid. Cl is the conjugate base of a strong+ - + -

acid. Cl has no basic (or acidic) properties.-

3 3 3 2 aCH NH º CH NH + H K = = 2.28 × 10+ + -11

3 3 3 2CH NH CH NH + H+ +

Initial 0.10 M 0 ~0

3 3x mol/L CH NH dissociates to reach equilibrium+


aK = 2.28 × 10 = (assuming x << 0.10)-11


x = [H ] = 1.5 × 10 M; pH = 5.82 Assumptions good.+ -6

b. NaCN ÷ Na + CN : CN is a weak base. Na has no acidic (or basic) properties.+ - - +

2 b CN + H O HCN + OH K = = 1.6 × 10- - -5


2x mol/L CN reacts with H O to reach equilibrium-


bK = 1.6 × 10 = -5

x = [OH ] = 8.9 × 10 M; pOH = 3.05; pH = 10.95 Assumptions good.- -4

2 2 2106. a. KNO ÷ K + NO : NO is a weak base. Ignore K .+ - - +

2 2 2 bNO + H O HNO + OH K = = 2.5 × 10- - -11


bK = 2.5 × 10 = . -11


b. NaOCl ÷ Na + OCl : OCl is a weak base. Ignore Na .+ - - +

2 bOCl + H O HOCl + OH K = = 2.9 × 10- - -7

Initial 0.45 M 0 ~0 Equil. 0.45 - x x x

bK = 2.9 × 10 = . -7


4 4 4 4 4 4c. NH ClO ÷ NH + ClO : NH is a weak acid. ClO is the conjugate base of a strong acid.+ - + -

4ClO has no basic (or acidic) properties.-

4 3 aNH NH + H K = = 5.6 × 10+ + -10

Initial 0.40 M 0 ~0


Equil. 0.40 - x x x

aK = 5.6 × 10 = . -10


107. All these salts contain Na , which has no acidic/basic properties, and a conjugate base of a weak acid+

(except for NaCl where Cl is a neutral species.). All conjugate bases of weak acids are weak bases-

b wsince K for these species is between K and 1. To identify the species, we will use the data given to

b bdetermine the K value for the weak conjugate base. From the K value and data in Table 14.2 of the

atext, we can identify the conjugate base present by calculating the K value for the weak acid. We willuse A as an abbreviation for the weak conjugate base.-

2 A + H O HA + OH - -

Initial 0.100 mol/1.00 L 0 ~0

2x mol/L A reacts with H O to reach equilibrium-


bK = ; From the problem, pH = 8.07:

pOH = 14.00 - 8.07 = 5.93; [OH ] = x = 10 = 1.2 × 10 M- -5.93 -6

b bK = = 1.4 × 10 = K value for the conjugate base of a weak acid.-11

a w b aThe K value for the weak acid equals K /K : K = = 7.1 × 10-4

aFrom Table 14.2 of the text, this K value is closest to HF. Therefore, the unknown salt is NaF.

108. BHCl ÷ BH + Cl ; Cl is the conjugate base of the strong acid HCl, so Cl has no acidic/basic+ - - -

aproperties. BH is a weak acid since it is the conjugate acid of a weak base, B. Determining the K+

value for BH :+

BH B + H + +

Initial 0.10 M 0 ~0x mol/L BH dissociates to reach equilibrium+



aK = ; From the problem, pH = 5.82:

a[H ] = x = 10 = 1.5 × 10 M; K = = 2.3 × 10+ -5.82 -6 -11

b w aK for the base, B = K \K = 1.0 × 10 /2.3 × 10 = 4.3 × 10 .-14 -11 -4

b 3 2 3 3From Table 14.3 of the text, this K value is closest to CH NH so the unknown salt is CH NH Cl.

2 6 a 3 2 w109. Major species present: Al(H O) (K = 1.4 × 10 ), NO (neutral) and H O (K = 1.0 × 10 );3+ -5 - -14

2 6Al(H O) is a stronger acid than water so it will be the dominant H producer.3+ +

2 6 2 5 Al(H O) Al(H O) (OH) + H 3+ 2+ +


2 6x mol/L Al(H O) dissociates to reach equilibrium3+


aK = 1.4 × 10 = -5

x = 8.4 × 10 M = [H ]; pH = -log (8.4 × 10 ) = 3.08; Assumptions good.-4 + -4

2 6 a 2 w 2 6110. Major species: Co(H O) (K = 1.0 × 10 ), Cl (neutral) and H O (K = 1.0 × 10 ); Co(H O)3+ -5 - -14 3+

will determine the pH since it is a stronger acid than water. Solving the weak acid problem in the usualmanner:

2 6 2 5 a Co(H O) Co(H O) (OH) + H K = 1.0 × 10 3+ 2+ + -5


aK = 1.0 × 10 = , x = [H ] = 1.0 × 10 M-5 + -3

pH = -log (1.0 × 10 ) = 3.00; Assumptions good.-3

111. Reference Table 14.6 of the text and the solution to Exercise 14.99 for some generalizations on acid-base properties of salts.

3 3a. NaNO ÷ Na + NO neutral; Neither species has any acidic/basic properties.+ -

2 2 2b. NaNO ÷ Na + NO basic; NO is a weak base and Na has no effect on pH.+ - - +


2 2 2 bNO + H O º HNO + OH K = = 2.5 × 10- - -11

5 5 4 5 5 4 5 5 4c. C H NHClO ÷ C H NH + C1O acidic; C H NH is a weak acid and ClO has no + - + -

effect on pH.

5 5 5 5 aC H NH º H + C H N K = = 5.9 × 10+ + -6

4 2 4 2 4 a 2d. NH NO ÷ NH + NO acidic; NH is a weak acid (K = 5.6 × 10 ) and NO is a weak base+ - + -10 -

b(K = 2.5 × 10 ). Since > , then the solution is acidic.-11

4 3 a 2 2 2 bNH º H + NH K = 5.6 × 10 ; NO + H O º HNO + OH K = 2.5 × 10+ + -10 - - -11

e. KOCl ÷ K + OCl basic; OCl is a weak base and K has no effect on pH.+ - - +

2 bOCl + H O º HOCl + OH K = = 2.9 × 10- - -7

4 4 4f. NH OCl ÷ NH + OCl basic; NH is a weak acid and OCl is a weak base. Since+ - + -

, then the solution is basic.

4 3 a 2 bNH º NH + H K = 5.6 × 10 ; OCl + H O º HOCl + OH K = 2.9 × 10+ + -10 - - -7

112. a. KCl ÷ K + Cl neutral; K and Cl have no effect on pH.+ - + -

4 2 3 2 4 2 3 2 4 2 3 2b. NH C H O ÷ NH + C H O neutral; NH is a weak acid and C H O is a weak base. Since+ - + -

, pH = 7.00.

4 3 aNH º NH + H K = = 5.6 × 10+ + -10

2 3 2 2 2 3 2 bC H O + H O º HC H O + OH K = = 5.6 × 10- - -10

3 3 3 3 3 3c. CH NH Cl ÷ CH NH + Cl acidic; CH NH is a weak acid and Cl has no effect + - + -

on pH.

3 3 3 2 aCH NH º H + CH NH K = = 2.28 × 10+ + -11

d. KF ÷ K + F basic; F is a weak base and K has no effect on pH.+ - - +

2 bF + H O º HF + OH K = = 1.4 × 10- - -11


4 4 4e. NH F ÷ NH + F acidic; NH is a weak acid and F is a weak base. Since ,+ - + -

then the solution is acidic.

4 3 a 2 bNH º H + NH K = 5.6 × 10 ; F + H O º HF + OH K = 1.4 × 10+ + -10 - - -11

3 3 3 3 3 3f. CH NH CN ÷ CH NH + CN basic; CH NH is a weak acid and CN is a weak base. Since+ - + -

, the solution is basic.

3 3 3 2 aCH NH º H + CH NH K = 2.28 × 10+ + -11

2 bCN + H O º HCN + OH K = = 1.6 × 10- - -5

Relationships Between Structure and Strengths of Acids and Bases

3 3113. a. HIO < HBrO ; As the electronegativity of the central atom increases, acid strength increases.

2 3b. HNO < HNO ; As the number of oxygen atoms attached to the central nitrogen atom increases,acid strength increases.

c. HOI < HOCl; Same reasoning as in a.

3 3 3 4d. H PO < H PO ; Same reasoning as in b.

3 3 3114. a. BrO < IO ; These are the conjugate bases of the acids in Exercise 14.113a. Since HBrO is the- -

3 3 3stronger acid, the conjugate base of HBrO (BrO ) will be the weaker base. IO will be the- -

3stronger base since HIO is the weaker acid.

3 2b. NO < NO ; These are the conjugate bases of the acids in Exercise 14.113b. Conjugate base- -

strength is inversely related to acid strength.

c. OCl < OI . These are the conjugate bases of the acids in Exercise 14.113c.- -

2 2 2115. a. H O < H S < H Se; As the strength of the H S X bond decreases, acid strength increases.

3 2 2 2 2 2 3 2b. CH CO H < FCH CO H < F CHCO H < F CCO H; As the electronegativity of neighboringatoms increases, acid strength increases.

4 3c. NH < HONH ; Same reason as in b.+ +

4 4d. NH < PH ; Same reason as in a.+ +

116. In general, the stronger the acid, the weaker the conjugate base.

a. SeH < SH < OH ; These are the conjugate bases of the acids in Exercise 14.115a. The ordering- - -

of the base strength is the opposite of the acids.


3 3b. PH < NH (See Exercise 14.115d.)

2 3c. HONH < NH (See Exercise 14.115c.)

117. In general, metal oxides form basic solutions in water and nonmetal oxides form acidic solutions inwater.

2 2 2a. basic; CaO(s) + H O(l) ÷ Ca(OH) (aq), Ca(OH) is a strong base.

2 2 2 3 2 3b. acidic; SO (g) + H O(l) ÷ H SO (aq), H SO is a weak diprotic acid.

2 2c. acidic; Cl O(g) + H O(l) ÷ 2 HOCl(aq), HOCl is a weak acid.

2 2118 a. basic; Li O(s) + H O(l) ÷ 2 LiOH(aq), LiOH is a strong base.

2 2 2 3 2 3b. acidic; CO (g) + H O(l) ÷ H CO (aq), H CO is a weak diprotic acid.

2 2 2c. basic; SrO(s) + H O(l) ÷ Sr(OH) (aq), Sr(OH) is a strong base.

Lewis Acids and Bases

119. A Lewis base is an electron pair donor, and a Lewis acid is an electron pair acceptor.

3 2 3 3a. B(OH) , acid; H O, base b. Ag , acid; NH , base c. BF , acid; F , base+ -

2 2 2120. a. Fe , acid; H O, base b. H O, acid; CN , base c. HgI , acid; I , base3+ - -

3 2121. Al(OH) (s) + 3 H (aq) ÷ Al (aq) + 3 H O(l) (Brønsted-Lowry base, H acceptor)+ 3+ +

3 4Al(OH) (s) + OH (aq) ÷ Al(OH) (aq) (Lewis acid, electron pair acceptor)- -

2 2122. Zn(OH) (s) + 2 H (aq) ÷ Zn (aq) + 2 H O(l) (Brønsted-Lowry base)+ 2+

2 4Zn(OH) (s) + 2 OH (aq) ÷ Zn(OH) (aq) (Lewis acid)- 2-

123. Fe should be the stronger Lewis acid. Fe is smaller and has a greater positive charge. Because of3+ 3+

this, Fe will be more strongly attracted to lone pairs of electrons as compared to Fe .3+ 2+

124. The Lewis structures for the reactants and products are:

2 2In this reaction, H O donates a pair of electrons to carbon in CO , which is followed by a proton shift

2 3 2 2to form H CO . H O is the Lewis base, and CO is the Lewis acid.


Additional Exercises

125. At pH = 2.000, [H ] = 10 = 1.00 × 10 M; At pH = 4.000, [H ] = 10 = 1.00 × 10 M+ -2.000 -2 + -4.000 -4

mol H present = 0.0100 L × = 1.00 × 10 mol H+ -4 +

Let V = total volume of solution at pH = 4.000: 1.00 ×10 mol/L = , V = 1.00 L-4

Volume of water added = 1.00 L - 0.0100 L = 0.99 L = 990 mL

126. 2.48 g TlOH × = 1.12 × 10 mol; TlOH is a strong base, so [OH ] = 1.12 × 10 M.-2 - -2

pOH = -log[OH ] = 1.951; pH = 14.000 - pOH = 12.049-

127. The light bulb is bright because a strong electrolyte is present, i.e., a solute is present that dissolvesto produce a lot of ions in solution. The pH meter value of 4.6 indicates that a weak acid is present.(If a strong acid were present, the pH would be close to zero.) Of the possible substances, only HCl

4(strong acid), NaOH (strong base) and NH Cl are strong electrolytes. Of these three substances, only

4NH Cl contains a weak acid (the HCl solution would have a pH close to zero and the NaOH solution

4 4would have a pH close to 14.0). NH Cl dissociates into NH and Cl ions when dissolved in water.+ -

4Cl is the conjugate base of a strong acid, so it has no basic (or acidic properties) in water. NH ,- +

3 4however, is the conjugate acid of the weak base NH , so NH is a weak acid and would produce a+

solution with a pH = 4.6 when the concentration is ~1 M.

6 5 2128. HBz H + Bz HBz = C H CO H+ -

oInitial C ~0 0 C = [HBz] = concentration of HBz thatdissolves to give saturated solution.

x mol/L HBz dissociates to reach equilibriumChange -x +x +xEquil. C - x x x

aK = = 6.4 × 10 = , where x = [H ]-5 +

6.4 × 10 = ; pH = 2.80; [H ] = 10 = 1.6 × 10 M-5 + -2.80 -3

C - 1.6 × 10 = = 4.0 × 10 , C = 4.0 × 10 + 1.6 × 10 = 4.2 × 10 M-3 -2 -2 -3 -2

6 5 2The molar solubility of C H CO H is 4.2 × 10 mol/L.-2


2 6 6 6129. For H C H O . = 7.9 × 10 and = 1.6 × 10 . Since , then the amount of H-5 -12 +

produced by the reaction will be negligible.

2 6 6 6 o[H C H O ] = = 0.0142 M

2 6 6 6 6 6 6H C H O (aq) HC H O (aq) + H (aq) = 7.9 × 10- + -5


= 7.9 × 10 = , x = 1.1 × 10 ; Assumption fails the 5% rule.-5 -3

Solving by the method of successive approximations:

7.9 × 10 = , x = 1.0 × 10 M (consistent answer)-5 -3

Since H produced by the reaction will be negligible, [H ] = 1.0 × 10 and pH = 3.00.+ + -3

o 2 4130. [H ] = 1.0 × 10 + 1.0 × 10 = 2.0 × 10 M from strong acids HCl and H SO .+ -2 -2 -2

4 a aHSO is a good weak acid (K = 0.012). However, HCN is a poor weak acid (K = 6.2 × 10 ) and- -10

4can be ignored. Calculating the H contribution from HSO :+ -

4 4 aHSO H + SO K = 0.012- + 2-

Initial 0.010 M 0.020 M 0Equil. 0.010 - x 0.020 + x x

aK = = 0.012 . , x = 0.0060; Assumption poor (60% error).

Using the quadratic formula: x + 0.032 x - 1.2 × 10 = 0, x = 3.4 × 10 M2 -4 -3

[H ] = 0.020 + x = 0.020 + 3.4 × 10 = 0.023 M; pH = 1.64+ -3

2 2 2 2131. For this problem we will abbreviate CH =CHCO H as Hacr and CH =CHCO as acr .- -

a. Solving the weak acid problem:

aHacr H + acr K = 5.6 × 10+ - -5


= 5.6 × 10 . , x = [H ] = 2.4 × 10 M; pH = 2.62; Assumptions good.-5 + -3


b. % dissociation = × 100 = × 100 = 2.4%

c. acr is a weak base and the major source of OH in this solution.- -

bK = 2 acr + H O Hacr + OH- -

bInitial 0.050 M 0 ~0 K = 1.8 × 10-10

Equil. 0.050 - x x x

bK = = 1.8 × 10 = -10


7 4 2 b 7 4 2132. From the pH, C H ClO is a weak base. Use the weak base data to determine K for C H ClO- -

(which we will abbreviate as CB ).-

2 CB + H O HCB + OH- -


Since pH = 8.65, pOH = 5.35 and [OH ] = 10 = 4.5 × 10 M = x.- -5.35 -6

bK = = 1.0 × 10-10

Since CB is a weak base, HCB, chlorobenzoic acid, is a weak acid. Solving the weak acid -

problem:

HCB H + CB+ -


aK = = 1.0 × 10 = . -4

x = [H ] = 4.5 × 10 M; pH = 2.35 Assumptions good.+ -3

2 6 2 2 5 3133. a. Fe(H O) + H O Fe(H O) (OH) + H O3+ 2+ +

Initial 0.10 M 0 ~0 Equil. 0.10 - x x x


aK = = 6.0 × 10 = -3

x = 2.4 × 10 ; Assumption is poor (x is 24% of 0.10). Using successive approximations:-2

= 6.0 × 10 , x = 0.021-3

= 6.0 × 10 , x = 0.022; = 6.0 × 10 , x = 0.022-3 -3

x = [H ] = 0.022 M; pH = 1.66+

b. Because of the lower charge, Fe (aq) will not be as strong an acid as Fe (aq). A solution of2+ 3+

iron(II) nitrate will be less acidic (have a higher pH) than a solution with the same concentrationof iron(III) nitrate.

134. See generalizations in Exercise 14.99.

aa. HI: strong acid; HF: weak acid (K = 7.2 × 10 )-4

bNaF: F is the conjugate base of the weak acid HF so F is a weak base. The K value for - -

w a, HFF = K /K = 1.4 × 10 . Na has no acidic or basic properties.- -11 +

NaI: neutral (pH = 7.0); Na and I have no acidic/basic properties.+ -

To place in order of increasing pH, we place the compounds from most acidic (lowestpH) to most basic (highest pH). Increasing pH: HI < HF < NaI < NaF.

4 4 ab. NH Br: NH is a weak acid (K = 5.6 × 10 ) and Br is a neutral species.+ -10 -

HBr: strong acidKBr: neutral; K and Br have no acidic/basic properties+ -

3 bNH : weak base, K = 1.8 × 10-5

4 3Increasing pH: HBr < NH Br < KBr < NHmost mostacidic basic

6 5 3 3 6 5 3c. C H NH NO : C H NH is a weak acid = 1.0 × 10 /3.8 × 10 = 2.6 × 10 )+ -14 -10 -5

3and NO is a neutral species.-

3 3NaNO : neutral; Na and NO have no acidic/basic properties.+ -

NaOH: strong base

6 5 aHOC H : weak acid (K = 1.6 × 10 )-10

6 5 6 5KOC H : OC H is a weak base ( = 6.3 × 10 ) and K is a- -5 +

neutral species.

6 5 2 bC H NH : weak base (K = 3.8 × 10 )-10


3HNO : strong acid

This is a little more difficult than the previous parts of this problem because two weak acids and

6 5 3two weak bases are present. Between the weak acids, C H NH is a stronger weak acid than+

6 5 a 6 5 3 a 6 5HOC H since the K value for C H NH is larger than the K value for HOC H . Between the+

b 6 5 b 6 5 2two weak bases, since the K value for OC H is larger than the K value for C H NH , then-

6 5 6 5 2OC H is a stronger weak base than C H NH .-

3 6 5 3 3 6 5 3 6 5 2 6 5 Increasing pH: HNO < C H NH NO < HOC H < NaNO < C H NH < KOC H < NaOH most most acidic basic

4 4135. Solution is acidic from HSO º H + SO . Solving the weak acid problem:- + 2-

4 4 aHSO H + SO K = 1.2 × 10- + 2- -2


1.2 × 10 = , x = 0.035-2

Assumption is not good (x is 35% of 0.10). Using successive approximations:

= 1.2 × 10 , x = 0.028-2

= 1.2 × 10 , x = 0.029; = 1.2 × 10 , x = 0.029-2 -2

x = [H ] = 0.029 M; pH = 1.54 +

136. The relevant reactions are:

2 3 3 3 3H CO º H + HCO = 4.3 × 10 ; HCO º H + CO = 5.6 × 10+ - -7 - + 2- -11

Initially, we deal only with the first reaction (since >> ) and then let these results controlvalues of concentrations in the second reaction.

2 3 3H CO H + HCO+ -


= 4.3 × 10 = -7


3x = 6.6 × 10 M = [H ] = [HCO ] Assumptions good.-5 + -

3 3 HCO H + CO- + 2-

Initial 6.6 × 10 M 6.6 × 10 M 0-5 -5

Equil. 6.6 × 10 - y 6.6 × 10 + y y-5 -5

3If y is small, then [H ] = [HCO ] and = 5.6 × 10 = . y+ - -11

3y = [CO ] = 5.6 × 10 M Assumptions good.2- -11

The amount of H from the second dissociation is 5.6 × 10 M or:+ -11

× 100 = 8.5 × 10 %-5

This result justifies our treating the equilibria separately. If the second dissociation contributed asignificant amount of H , then we would have to treat both equilibria simultaneously. The reaction+

3that occurs when acid is added to a solution of HCO is:-

3 2 3 2 2HCO (aq) + H (aq) ÷ H CO (aq) ÷ H O(l) + CO (g)- +

2 2 3The bubbles are CO (g) and are formed by the breakdown of unstable H CO molecules. We should

2 2 2write H O(l) + CO (aq) or CO (aq) for what we call carbonic acid. It is for convenience, however,

2 3that we write H CO (aq).

2 2 4137. a. In the lungs, there is a lot of O and the equilibrium favors Hb(O ) . In the cells, there is a

2 4deficiency of O , and the equilibrium favors HbH .4+

2 2 2 3 2b. CO is a weak acid, CO + H O º HCO + H . Removing CO essentially decreases H . - + +

2 4 2Hb(O ) is then favored and O is not released by hemoglobin in the cells. Breathing into a paper

2bag increases CO in the blood, thus increasing H , which shifts the reaction left.+

2c. CO builds up in the blood and it becomes too acidic, driving the equilibrium to the left.

2Hemoglobin can't bind O as strongly in the lungs. Bicarbonate ion acts as a base in water andneutralizes the excess acidity.

3 3 4 2138. a. NH + H O º NH + H O+ +

eqK = = 1.8 × 109

2 3 2 2 eqb. NO + H O º HNO + H O K = = 2.5 × 10- + 3

4 3 2 eqc. NH + OH º NH + H O K = = 5.6 × 10+ - 4

2 2 2d. HNO + OH º H O + NO- -


eqK = = 4.0 × 1010

2 3 3 3 3139. a. H SO b. HClO c. H PO

NaOH and KOH are soluble ionic compounds composed of Na and K cations and OH anions. All+ + -

soluble ionic compounds dissolve to form the ions from which they are formed. In oxyacids, thecompounds are all covalent compounds in which electrons are shared to form bonds (unlike ioniccompounds). When these compounds are dissolved in water, the covalent bond between oxygen andhydrogen breaks to form H ions.+

Challenge Problems

140. The pH of this solution is not 8.00 because water will donate a significant amount of H from the+

autoionization of water. The pertinent reactions are:

2 wH O º H + OH K = [H ] [OH ] = 1.0 × 10+ - + - -14

aHCl ÷ H + Cl K is very large, so we assume that only the forward reaction occurs.+ -

In any solution, the overall net positive charge must equal the overall net negative charge (called thecharge balance). For this problem:

[positive charge] = [negative charge], so [H ] = [OH ] + [Cl ]+ - -

w wFrom K , [OH ] = K /[H ], and from 1.0 × 10 M HCl, [Cl ] = 1.0 × 10 M. Substituting into the- + -8 - -8

charge balance equation:

[H ] = + 1.0 × 10 , [H ] - 1.0 × 10 [H ] - 1.0 × 10 = 0+ -8 + 2 -8 + -14

Using the quadratic formula to solve:

[H ] = , [H ] = 1.1 × 10 M+ + -7

pH = -log (1.1 × 10 ) = 6.96-7

141. Since this is a very dilute solution of NaOH, we must worry about the amount of OH donated from-

the autoionization of water.

NaOH ÷ Na + OH+ -

2 wH O º H + OH K = [H ] [OH ] = 1.0 × 10+ - + - -14

This solution, like all solutions, must be charge balanced, that is [positive charge] = [negative charge].For this problem, the charge balance equation is:


[Na ] + [H ] = [OH ], where [Na ] = 1.0 × 10 M and [H ] = + + - + -7 +

Substituting into the charge balance equation:

1.0 × 10 + = [OH ], [OH ] - 1.0 × 10 [OH ] - 1.0 × 10 = 0-7 - - 2 -7 - -14

Using the quadratic formula to solve:

[OH ] = -

[OH ] = 1.6 × 10 M; pOH = -log (1.6 × 10 ) = 6.80; pH = 7.20- -7 -7

a142. HA H + A K = 1.00 × 10+ - -6

oInitial C ~0 0 C = [HA] ; For pH = 4.000,Equil. C - 1.00 × 10 1.00 × 10 1.00 × 10 x = [H ] = 1.00 × 10 M-4 -4 -4 + -4

aK = = 1.00 × 10 ; Solving: C = 0.0101 M-6

The solution initially contains 50.0 × 10 L × 0.0101 mol/L = 5.05 × 10 mol HA. We then dilute-3 -4

to a total volume, V, in liters. The resulting pH = 5.000, so [H ] = 1.00 × 10 . In the typical weak+ -5

acid problem, x = [H ], so:+

HA H + A+ -

Initial 5.05 × 10 mol/V ~0 0-4

Equil. 5.05 × 10 /V - 1.00 × 10 1.00 × 10 1.00 × 10-4 -5 -5 -5

aK = = 1.00 × 10 , 1.00 × 10 = 5.05 × 10 /V - 1.00 × 10-6 -4 -4 -5

V = 4.59 L; 50.0 mL are present initially, so we need to add 4540 mL of water.

a143. HBrO H + BrO K = 2 × 10+ - -9

Initial 1.0 × 10 M ~0 0-6

x mol/L HBrO dissociates to reach equilibriumChange -x +x +xEquil. 1.0 × 10 - x x x-6


aK = 2 × 10 = ; x = [H ] = 4 × 10 M; pH = 7.4-9 + -8

Let’s check the assumptions. This answer is impossible! We can't add a small amount of an acid toa neutral solution and get a basic solution. The highest possible pH for an acid in water is 7.0. In thecorrect solution, we would have to take into account the autoionization of water.

2 5 5 a w b 5 5144. Major species present are H O, C H NH (K = K /K (C H N) = 1.0 × 10 /1.7 × 10 = + -14 -9

b w a5.9 × 10 ) and F (K = K /K (HF) = 1.0 × 10 /7.2 × 10 = 1.4 × 10 ). The reaction to consider-6 - -14 -4 -11

5 5is the best acid present (C H NH ) reacting with the best base present (F ). Solving for the equilibrium+ -

concentrations:

5 5 5 5C H NH (aq) + F (aq) C H N(aq) + HF(aq)+ -

Initial 0.200 M 0.200 M 0 0Change -x -x +x +xEquil. 0.200 - x 0.200 - x x x

K = × = 5.9 × 10 (1/7.2 × 10 ) = 8.2 × 10-6 -4 -3

K = = 8.2 × 10 = ; Taking the square root of both sides:-3

0.091 = x = 0.018 - 0.091 x, x = 0.016 M

5 5From the setup to the problem, x = [C H N] = [HF] = 0.016 M and 0.200 - x = 0.200 - 0.016 = 0.184

5 5 a 5 5M = [C H NH ] = [F ]. To solve for the [H ], we can use either the K equilibrium for C H NH or+ - + +

a 5 5the K equilibrium for HF. Using C H NH data:+

= 5.9 × 10 = , [H ] = 6.8 × 10 M-6 + -5

pH = -log(6.8 × 10 ) = 4.17-5

a bAs one would expect, since the K for the weak acid is larger than the K for the weak base, then asolution of this salt should be acidic.

3 3145. Since NH is so concentrated, we need to calculate the OH contribution from the weak base NH .-

3 2 4 b NH + H O NH + OH K = 1.8 × 10+ - -5

Initial 15.0 M 0 0.0100 M (Assume no volume change.)Equil. 15.0 - x x 0.0100 + x

bK = 1.8 × 10 = , x = 0.027; Assumption is horrible (x is 270% -5

of 0.0100).

Using the quadratic formula:

1.8 × 10 (15.0 - x) = 0.0100 x + x , x + 0.0100 x - 2.7 × 10 = 0-5 2 2 -4


x = 1.2 × 10 , [OH ] = 1.2 × 10 + 0.0100 = 0.022 M-2 - -2

146. Molar mass = = 125 g/mol

o [HA] = = 0.120 M; pH = 1.80, [H ] = 10 = 1.6 × 10 M+ -1.80 -2

HA H + A+ -

Equil. 0.120 - x x x x = [H ] = 1.6 × 10 M+ -2

aK = = 2.5 × 10-3

4 4 a 4147. PO is the conjugate base of HPO . The K value for HPO is = 4.8 × 10 .3- 2- 2- -13

4 2 4 bPO (aq) + H O(l) HPO (aq) + OH (aq) K = = 0.0213- 2- -

4 2 4HPO is the conjugate base of H PO ( = 6.2 × 10 ).2- - -8

4 2 2 4 bHPO + H O H PO + OH K = = 1.6 × 102- - - -7

2 4 3 4H PO is the conjugate base of H PO ( = 7.5 × 10 ).- -3

2 4 2 3 4 bH PO + H O H PO + OH K = = 1.3 × 10- - -12

b 4 4From the K values, PO is the strongest base. This is expected since PO is the conjugate base of3- 3-

4the weakest acid (HPO ).2-

3 3 2 3 3148. a. HCO + HCO H CO + CO- - 2-

eqK = = 1.3 × 10-4

2 3 3b. [H CO ] = [CO ] since the reaction in part a is the principle equilibrium reaction.2-

2 3 3 eqc. H CO 2 H + CO K = + 2-

2 3 3Since, [H CO ] = [CO ] from part b, then [H ] = .2- + 2

[H ] = ( ) or pH = + 1/2


d. [H ] = [(4.3 × 10 ) × (5.6 × 10 )] , [H ] = 4.9 × 10 M; pH = 8.31+ -7 -11 1/2 + -9

149. Molality = m = = 2.00 × 10 mol/kg . 2.00 × 10 mol/L (dilute solution)-3 -3

f f)T = iK m, 0.0056°C = i(1.86°C/molal) (2.00 × 10 molal), i = 1.5-3

If i = 1.0, % dissociation = 0% and if i = 2.0, % dissociation = 100%. Since i = 1.5, the weak acidis 50.% dissociated.

aHA H + A K = + -

Since the weak acid is 50.% dissociated, then:

o[H ] = [A ] = [HA] × 0.50 = 2.00 × 10 M × 0.50 = 1.0 × 10 M+ - -3 -3

o[HA] = [HA] - amount HA reacted = 2.00 × 10 M - 1.0 × 10 M = 1.0 × 10 M-3 -3 -3

aK = = 1.0 × 10-3

4 2 4 3150. a. Assuming no ion association between SO (aq) and Fe (aq), then i = 5 for Fe (SO ) .2- 3+

B = iMRT = 5(0.0500 mol/L)(0.08206 L atm K mol )(298 K) = 6.11 atm-1 -1

2 4 3 4b. Fe (SO ) (aq) ÷ 2 Fe (aq) + 3 SO (aq)3+ 2-

4Under ideal circumstances, 2/5 of B calculated above results from Fe and 3/5 results from SO .3+ 2-

4 4The contribution to B from SO is 3/5 × 6.11 atm = 3.67 atm. Since SO is assumed unchanged2- 2-

4in solution, the SO contribution in the actual solution will also be 3.67 atm. The contribution2-

2 6to the actual B from the Fe(H O) dissociation reaction is 6.73 - 3.67 = 3.06 atm.3+

2 6The initial concentration of Fe(H O) is 2(0.0500) = 0.100 M. The set-up for the weak acid2+

problem is:

2 6 2 5 aFe(H O) H + Fe(OH)(H O) K = 3+ + 2+


2 6x mol/L of Fe(H O) reacts to reach equilibrium3+

Equil. 0.100 - x x x

B = iMRT; Total ion concentration = iM = = 0.125 M

0.125 M = 0.100 - x + x + x = 0.100 + x, x = 0.025 M

a aK = , K = 8.3 × 10-3

CHAPTER FOURTEEN ACIDS AND BASES...346 CHAPTER 14 ACIDS AND BASES bases. The nitrogen atom has an...

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Transcript of CHAPTER FOURTEEN ACIDS AND BASES...346 CHAPTER 14 ACIDS AND BASES bases. The nitrogen atom has an...