AP Chemistry

AP Chemistry

Ch. 14 -Acids, Bases, and Acid-Base Equilibrium

Properties of Acids

• pH below 7• Conduct electricity in solution• React with to form Products

– Metals to form H2

– Carbonates to form CO2, H20, and a salt

– Bicarbonates to form CO2, and H20

– A Base to form H20 and a salt

Properties of Bases

• pH above 7

• Turns red litmus paper to blue Turns red litmus paper to blue

• ““BBasic asic BBlue”lue”

• Conduct electricity in solution

• React with to form Products– Metal ions to form Precipitates

– Bicarbonates to form CO3, and H20

– An acid to form H20 and a salt

Definitions of Acids & Bases• Arrhenius Definition of Acids and Bases:

• Most elementary of ideas about acids

• a) Acids produce H+ ions in solutions

• b) Bases produce (OH)- ions in solutions

Strong Acids

• HI – hydroiodic 7 you need

• HBr-hydrobromic to remember.

• HCl - hydrochloric

• HClO4 - perchloric

• H2SO4 - sulfuric

• HClO3 - chloric

• HNO3 - nitric

Acid and Base Strength

• Strong acids are completely dissociated in water.– Their conjugate bases are

quite weak.

• Weak acids only dissociate partially in water.– Their conjugate bases are

weak bases.

Strong Bases

• LiOH – lithium hydroxide

• NaOH Sodium Hydroxide

• KOH Potassium Hydroxide

• Ca(OH)2 Calcium Hydroxide

• Sr(OH)2 Strontium Hydroxide

• Ba(OH)2 Barium Hydroxide

6 you need to remember

Problems with Arrhenius• The Arrhenius definitions of acids

and bases did not properly explain why other substances are acids or bases

• Example) NH3

Bronsted-Lowry Definition

• These two scientists devise better definitions for acids and bases which would encompass more substances

• Bronsted-Lowry Acid: A proton donor• Bronsted-Lowry Base: A proton acceptor

More About Bronsted-Lowry• A proton is simply a H+ ion

• The hydrogen in substances is described as being ionizable

• This ionizable hydrogen is attracted to a center of negative charge (lone pairs of electrons)

More About Bronsted-Lowry• HCl + H2O H3O+ + Cl-

H Cl

H

HO

Cl-

H

HO+H

More About Bronsted-Lowry• 1) Show the reaction between HCl

and NH3 and clearly identify the acid and the base

• 2) Identify the acid and the base when NH3 is placed in water

Conjugate Acids and Bases• When an acid-base reactions occurs, a

conjugate acid and base is formed• Conjugate acids are the original base

plus a hydrogen ion (ie it is the acid on the product side of the equation)

Conjugate Acids and Bases

• Conjugate bases are the original acid minus a hydrogen ion (ie it is the base on the product side of the equation)

Conjugate Acids and Bases:

• Reactions between acids and bases always yield their conjugate bases and acids.

Identify the Bronsted-Lowry Acids and Bases & their conjugates

• 1) H2S + NH3 NH4+ + HS-

Acid base con. Base con. base

• 2) OH- + H2PO4- H2O + HPO4

2

base acid con. acid con. base

Significance of Conjugate Acid-Bases• The stronger an acid, the weaker is its

conjugate base

• The stronger a base, the weaker is its conjugate acid

• Acid-Base reactions favor the direction of the stronger member to the weaker member of each pair

Acids & Base DefinitionsAcids & Base Definitions

Lewis acid - a substance that Lewis acid - a substance that accepts an electron pairaccepts an electron pair

Lewis base - a substance Lewis base - a substance that donates an electron that donates an electron pairpair

Definition #3 – Lewis Definition #3 – Lewis

Formation ofFormation of hydronium ion hydronium ion is also an is also an excellent example.excellent example.

Lewis Acids & BasesLewis Acids & Bases

•Electron pair of the new O-H bond Electron pair of the new O-H bond originates on the Lewis base.originates on the Lewis base.

HH

H

BASE

••••••

O—HO—H

H+

ACID

Lewis Acid/Base ReactionLewis Acid/Base Reaction

AP Practice

• In the lab H2(g) can be produced by adding which of the following to 1.0M HCl(aq)?

• I. 1 M NH3 (aq)

• II. Zn (s)

• III. NaHCO3 (s)

A. I only C. III only E. I, II, and III

B. II only D. I and II only

AP Practice• In liquid ammonia, the reaction represented

below occurs.

2NH3 NH4+ + NH2

-

In the reaction, NH4+ acts as:

• A. a catalyst

• B. Both an acid and a base

• C. the conjugate acid of NH3

• D. The reducing agent

• E. the oxidizing agent

AP Practice #4

• Write the balance net ionic equation for :

A 0.1M nitrous acid solution is added to the same volume of a 0.1M sodium hydroxide solution.

HNO2 + OH- NO2- + H2O

AP Practice #4

• Write the net ionic equation for:

Hydrogen iodide gas is bubbled into a solution of lithium carbonate.

2 HI + CO3-2 2 I- + H20 + CO2

AP Practice #4

• Write the balanced net ionic equation for: Concentrated hydrochloric acid is added to a solution of sodium sulfide.

2 H+ + S-2 H2S

AP Practice #4

• Write the balanced net ionic equation for: A solution of ethanoic (acetic) acid is added to a solution of barium hydroxide:

HC2H3O2 + OH- H2O + C2H3O2 -

Strength of Acids And Bases

Strong and Weak Acids/BasesStrong and Weak Acids/Bases

The strength of an acid (or base) is determined by the amount of IONIZATION.

The strength of an acid (or base) is determined by the amount of IONIZATION.

• Weak acidsWeak acids are much less than 100% ionized in water. are much less than 100% ionized in water.

*One of the best known is acetic acid = CH*One of the best known is acetic acid = CH33COCO22HH


Strong Acid Weak Acid

15.4

• Strong Base:Strong Base: 100% dissociated in water. Have 100% dissociated in water. Have very weak conjugate acids.very weak conjugate acids.

NaOH (aq) ---> NaNaOH (aq) ---> Na+ + (aq) + OH(aq) + OH- - (aq)(aq)

Other common strong Other common strong bases include KOH andbases include KOH and Ca(OH)Ca(OH)22..

CaO (lime) + HCaO (lime) + H22O -->O -->

Ca(OH)Ca(OH)22 (slaked lime) (slaked lime)

CaOCaO


• Weak base:Weak base: less than 100% ionized in waterless than 100% ionized in water

One of the best known weak bases is ammoniaOne of the best known weak bases is ammonia

NHNH3 3 (aq) + H(aq) + H22O (l) O (l) NH NH44+ + (aq) + OH(aq) + OH- - (aq)(aq)

Strong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/Bases

Weak BasesWeak Bases

Significance of Acid/Base Strength

• When equilibrium is established, the side in which there are stronger acids/bases will shift toward the weaker sides (LeChatlier Principle)

• Thus the concentration of substances will favor the weaker members

Factors Affecting Acid Strength

• A) Binary Acids: The lower the bond dissociation energy, the easier the bond is broken.

• More likely that acid will donate a H ion.

• (Low BE = Stronger Acid (Weaker Conjugate Base)

• ***Low BE = Stronger Acid***

Molecular Structure and Acid Strength

H X H+ + X-

The stronger the bond

The weaker the acid

HF << HCl < HBr < HI

• Bond strength

• Polarity


• The larger the anion, the stronger that acid is (the easier the bond is broken)

• Acid strength increases going across the table right to left while

• {the greater the distribution in charge (polarity) the more likely the substance will lose H ion}


• Nonpolar covalent acids are weaker than polar covalent acids which are weaker than ionic acids

•Essentially, the greater the dipole in the acid, the more likely the acid is strong!

Molecular Structure and Acid Strength

Z O H Z O- + H+

- +

The O-H bond will be more polar and easier to break if:

• Z is very electronegative or

• Z is in a high oxidation state

AP Chemistry Quiz

• 1. Explain how to determine between a Bronsted-Lowry Acid and Base.

• 2. Why is HBr is a stronger acid than H2S?

Oxoacids: • B) Oxoacids: Contain hydrogen, oxygen,

and some other element (nonmetal). At least one H bonded to an O.

• The other element’s tendency to attract other electrons assists in determining the strength of the acid

Oxoacids:

• If the other element attracts electrons very strongly, electrons are withdrawn from Oxygen – Hydrogen bond. This weakens the O-H bond and results in stronger acids.

Molecular Structure and Acid Strength1. Oxoacids having different central atoms (Z) that are from the same group and that have the same oxidation number.

Acid strength increases with increasing electronegativity of Z

H O Cl O

O• •

• •

• •

• • ••

• •

• •

••••

H O Br O

O• •

• •

• •

• • ••

• •

• •

••••Cl is more electronegative than Br

HClO3 > HBrO3

• Adding more oxygen atoms that are added (H2SO4 and H2SO3) is the same as adding more electronegative elements.

• Since oxygen has a high electronegativity, this withdraws electrons from the O-H bond and results in stronger acids

Molecular Structure and Acid Strength2. Oxoacids having the same central atom (Z) but different numbers of attached groups.

Acid strength increases as the oxidation number of Z increases.

HClO4 > HClO3 > HClO2 > HClO

pH Scale and its calculations

pH Scale

• The pH Scale runs 0-14

• pH values < 7 are acidic

• pH values > 7 are basic (alkaline)

• pH values = 7, Neutral

How Do We Measure pH?

• For less accurate measurements, one can use– Litmus paper

• “Red” paper turns blue above ~pH = 8

• “Blue” paper turns red below ~pH = 5

– An indicator

How Do We Measure pH?

For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

Mathematical Determination of pH

• pH = - log [H+] OR

• - log [H3O+]

Mathematical Determination of pOH & Other Relationships

• pOH = - log [OH-]

• pH + pOH = 14

• [H+] * [OH-] = 1 x 10-14 = Kw

pKa and pKb Values

• pKa = - log Ka

• pKb = - log Kb

• Low values for pKa & pKb correspond to large values for Ka and Kb

Other “p” Scales

• The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions).

• Some similar examples are– pOH = −log [OH−]

– pKw = −log Kw

Homework #38

Kw= [H+] [OH-]

a. [H+] = = 6.7 × 10-15 M

pH=-log [H+] =14.1 Basic

b.[H+] = = 2.8 M; acidic

5.1

100.1

][

14

OH

K w

15

14

106.3

100.1

38 Continued

c. [H+] = = 1.0 × 10-7 M; neutral

d. [H+] = = 1.4 × 10-11M; basic

7

14

100.1

100.1

4

14

103.7

100.1

Number 44

pH pOH

• 9.63 14.00 – 9.63 = 4.37

• [H]

• 10-pH =[H+] = = 2.3 × 10-10M

• [OH]

• 10-pOH =[ OH] = = 4.3 × 10-5 M;

• acid/base/neutral

• BASIC

63.910

37.410

Calculating the pH of strong acids or bases

• Since strong acids and bases completely ionize, we assume that no reactant remains together.

• Focus on the major species present in solution.

• To help understand lets calculate the pH of a 1M HCl solution.

Stong acid pH

• Major species in solution:

• H+, Cl- H20 (does not contribute enough H)

• 1M [H+] and 1M of [Cl-] are in solution

• pH= -log [H+]

• So pH= -log[1]

• pH=0

•

Problem 48

• Strong acids are assumed to completely dissociate in water: HCl(aq) → H+(aq) + Cl (aq)

a. 0.10 M HCl solution gives 0.10 M H+ and 0.10 M Cl since HCl completely dissociates. pH = -log [H+] = -log (0.10) = 1.00

b. 5.0 M H+ is produced when 5.0 M HCl completely dissociates. pH = -log (5.0) = -0.70 (Negative pH values just indicate very concentrated acid solutions).

•

48 Cont.

• C. 1.0 × M H+ is produced when 1.0 × M HCl completely dissociates. This gives pH = 11.00. This is impossible!

• We dissolved an acid in water and got a basic pH. What we must consider in this problem is that water by itself donates 1.0 × 10-7 M H+.

• We can normally ignore the small amount of H+ from H2O except when we have 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.

Weak Acids

• Weak acids do not ionize 100% but instead establish an equilibrium or a Ka value.

• Still use products over reactants to calculate Ka.

• Must now focus on major species in a reaction and use ICE charts.

Dissociation Constants

• For a generalized acid dissociation,

the equilibrium expression would be

• This equilibrium constant is called the acid-dissociation constant, Ka.

[H3O+] [A−][HA]

Ka =

HA(aq) + H2O(l) A−(aq) + H3O+(aq)

Dissociation Constants

The greater the value of Ka, the stronger the acid.

Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid

You have 1.00 M HOAc. Calc. the equilibrium You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hconcs. of HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.

Step 1.Step 1. Define equilibrium conc. in ICE table.Define equilibrium conc. in ICE table.

[HOAc][HOAc] [H[H33OO++]] [OAc[OAc--]]

initialinitial

changechange

equilibequilib

1.001.00 00 001.001.00 00 00

-x-x +x+x +x+x-x-x +x+x +x+x

1.00-x1.00-x xx xx1.00-x1.00-x xx xx

Step 2.Step 2. Write KWrite Kaa expression and obtain expression and obtain

Ka value from a table.Ka value from a table.

You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, HYou have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H33OO++, ,

OAcOAc--, and the pH., and the pH.

Ka 1.8 x 10-5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

This is a quadratic. Solve using quadratic This is a quadratic. Solve using quadratic formula.formula.

or you can make an approximation if x is very or you can make an approximation if x is very small! (Rule of thumb: 10small! (Rule of thumb: 10-5-5 or smaller is ok this or smaller is ok this will usually satisfy the 5% rule)will usually satisfy the 5% rule)

or you can make an approximation if x is very or you can make an approximation if x is very small! (Rule of thumb: 10small! (Rule of thumb: 10-5-5 or smaller is ok this or smaller is ok this will usually satisfy the 5% rule)will usually satisfy the 5% rule)



Step 3.Step 3. Solve KSolve Kaa expression expression



Ka 1.8 x 10-5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

First assume x is very small because First assume x is very small because KKaa is so small. is so small.

Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00

Now we can more easily solve this Now we can more easily solve this approximate expression.approximate expression.

Step 3.Step 3. Solve KSolve Kaa approximateapproximate expressionexpression



Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00

x =x = [[HH33OO++] = [] = [OAcOAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M

pH = - log [pH = - log [HH33OO++] = -log (4.2 x 10] = -log (4.2 x 10-3-3) =) = 2.372.37



Calculate the pH of a 0.0010 M solution of formic Calculate the pH of a 0.0010 M solution of formic acid, HCOacid, HCO22H.H.

HCOHCO22H + HH + H22O O HCO HCO22-- + H + H33OO++

KKaa = 1.8 x 10 = 1.8 x 10-4-4

Approximate solutionApproximate solution

[H[H33OO++] = 4.2 x 10] = 4.2 x 10-4-4 M, M, pH = 3.37pH = 3.37

Exact SolutionExact Solution

[H[H33OO++] = [HCO] = [HCO22--] = 3.4 x 10] = 3.4 x 10-4-4 M M

[HCO[HCO22H] = 0.0010 - 3.4 x 10H] = 0.0010 - 3.4 x 10-4-4 = 0.0007 M = 0.0007 M

pH = 3.47 pH = 3.47

Problem #54

• a. HOC6H5 (Ka = 1.6 × 10-10 ) and H2O (Ka = Kw = 1.0 × 10-14 ) are the major species. The major equilibrium is the dissociation of HOC6H5. Solving the weak acid problem:

• OC6H5 ⇌ H+ + OC6H5

Initial 0.250 M ~0 0

Change -x → +x +x

Equil. 0.250 - x x x

Problem #54 continued

Ka = 1.6 × 10-10 = = =

(assuming x << 0.250)

• x = [H+] = 6.3 × 10-6 M

• Checking assumption: x is 2.5 × 10-3 % of 0.250, so assumption is valid by the 5% rule.

•

• pH = -log(6.3 × 10-6) = 5.20

]HHOC[

]HOC][H[

56

56

x

x

250.0

2

250.0

2x

54 (b)

• x = [H+] = 1.2 × 10-5 M;

• Checking assumption: x is 4.8 × 10-3 % of 0.250

• Assumptions good.

• pH = -log (1.2 × 10-5 ) = 4.92

58

• Major species: HIO3, H2O; Major source of H+: HIO3 (a weak acid, Ka = 0.17)

• HIO3 ⇌ H+ + IO3

• Initial 0.20 M ~0 0

• Change -x → +x +x

• Equil. 0.20 –x x x

•

58 Cont.

• Ka = 0.17 = = x = 0.18; Check assumption.

Assumption is horrible (x is 90% of 0.20).

• 0.17 =

• x2 = 0.17(0.20 – x)

• x2 + 0.17 x - 0.034 = 0

• x =

• x = 0.12 or -0.29 Only x = 0.12 makes sense.

x

x

20.0

2

20.0

2x

x

x

020.0

2

2

1406.017.0

)1(2

)]034.0)(1(4)17.0[(17.0 2/12

58 Answer

• x = 0.12 M = [H+];

• pH = -log (0.12) = 0.92

Equilibria Involving A Weak BaseEquilibria Involving A Weak Base

You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.

NHNH33 + H + H22O O NH NH44++ + OH + OH--

KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 1.Step 1. Define equilibrium concs. in ICE tableDefine equilibrium concs. in ICE table

[NH[NH33]] [NH[NH44++]] [OH[OH--]]

initialinitial

changechange

equilibequilib

0.0100.010 00 000.0100.010 00 00

-x-x +x+x +x+x-x-x +x+x +x+x

0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx

Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseYou have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.

NHNH33 + H + H22O O NH NH44++ + OH + OH--

KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression

Kb 1.8 x 10-5 = [NH4

+][OH- ][NH3 ]

= x2

0.010 - xKb 1.8 x 10-5 =

[NH4+][OH- ]

[NH3 ] =

x2

0.010 - x

Assume x is small, soAssume x is small, so x = [OHx = [OH--] = [NH] = [NH44

++] = 4.2 x 10] = 4.2 x 10-4-4 M M

and [NHand [NH33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10-4-4 ≈ 0.010 M ≈ 0.010 M

The approximation is validThe approximation is valid !!

Equilibria Involving A Weak Equilibria Involving A Weak BaseBase

Step 3.Step 3. Calculate pHCalculate pH

[OH[OH--] = 4.2 x 10] = 4.2 x 10-4-4 M M

so pOH = - log [OHso pOH = - log [OH--] = 3.37] = 3.37

Because pH + pOH = 14,Because pH + pOH = 14,

pH = 10.63pH = 10.63

Problem 88

• Kb = 1.3 × 10-3

• (C2H5)2NH + H2O (C⇌ 2H5)2NH2+ + OH

• Initial 0.050 M 0 ~0

• Change -x → +x +x

• Equil. 0.050 - x x x

88 cont.• Kb = 1.3 × 10-3 = = ≈

• x = 8.1 × 10-3;

• Assumption is bad (x is 16% of 0.20).

• Using Quadratic x= -b +/- (b2-4ac)1/2 / 2a

• 0=x2 + .0013x - 6.5 E-5

• X=.0067

• [OH] = x = .0067M;

• [H+] = Kw/[OH] = 1.49 × 10-12 M;

• pH = 11.83

]NH)HC[(

]OH][NH)HC[(

252

2252

x

x

050.0

2

050.0

2x

Calculating Ka from the pH

• The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature.

• We know that

[H3O+] [COO−][HCOOH]

Ka =


• The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature.

• To calculate Ka, we need the equilibrium concentrations of all three things.

• We can find [H3O+], which is the same as [HCOO−], from the pH.


pH = −log [H3O+]

2.38 = −log [H3O+]

−2.38 = log [H3O+]

10−2.38 = 10log [H3O+] = [H3O+]

4.2 10−3 = [H3O+] = [HCOO−]

Calculating Ka from pH

Now we can set up a table…

[HCOOH], M [H3O+], M [HCOO−], M

Initially 0.10 0 0

Change −4.2 10-3 +4.2 10-

3

+4.2 10−3

At Equilibrium

0.10 − 4.2 10−3

= 0.0958 = 0.10

4.2 10−3 4.2 10−3

Calculating Ka from pH

[4.2 10−3] [4.2 10−3][0.10]

Ka =

= 1.8 10−4

Homework #92

• Remember this is an equation for OH but they give you a pH. Think about how to get to OH.

Percent Ionization

Calculating Percent Ionization

• Percent Ionization = 100

• [H3O+]eq is obtained from your ICE chart

• In this example

[H3O+]eq = 4.2 10−3 M

[HCOOH]initial = 0.10 M

[H3O+]eq

[HA]initial

Calculating Percent Ionization

Percent Ionization = 1004.2 10−3

0.10

= 4.2%

Polyprotic Acids• Polyprotic acids are acids that have the

ability to ionize more than once

• Ex) H3PO4, H2SO4, H2CO3

• The Ka for the second and third ionization is much smaller than the first so we generally ignore their contributions to the H+ conc.

Polyprotic Acids• H3PO4 H3O+ + (H2PO4)-

• (H2PO4)- H3O+ + (HPO4)2-

• H(PO4)2- H3O+ + (PO4)3-

• Ka1 = 7.1 x 10-3, Ka2 = 6.3 x 10-8, Ka3 = 4.3 x 10-13

Problem #96

• 96. The reactions are:

• H3AsO4 H⇌ + + H2AsO4 = 5 × 10-3

• H2AsO4 H⇌ + + HAsO4

2 = 8 × 10-8

• HAsO42 H⇌ + + AsO4

3 = 6 × 10-10

96 cont.• We will deal with the reactions in order of

importance, beginning with the largest Ka, .

• H3AsO4 H⇌ + + H2AsO4

• Initial 0.20 M ~0 0

• Equil. 0.20 – x x x

• By using quadratic formula. X= 3 x 10-2

• [H+] = [H2AsO4] = 3 × 10-2 M;

• [H3AsO4] = 0.20 - 0.03 = 0.17 M

96 cont.• • Because Ka = = 8 × 10-8 is much smaller

than the value, very little of H2AsO4 (and HAsO4

2) dissociates compared to H3AsO4. Therefore, [H+] and [H2AsO4

] will not change significantly by the reaction. Using the previously calculated concentrations of H+ and H2AsO4

we can calculate the concentration of HAsO4

2:

•

]AsOH[

]HAsO[]H[

42

24

96.• 8 × 10-8 M =

• [HAsO42] = 8 × 10-8 M

• Assumption that the Ka2 reaction does not change [H+] and [HAsO4

] is good because it is very small compared to Ka1

• We repeat the process using Ka3 to get [AsO43].

• = =

• [AsO43] = 1.6 × 10-15≈ 2 × 10-15 M

2

24

2

103

]HAsO[)103(

][

][][2

4

34

HAsO

AsOH)108(

]AsO[)103(8

34

2

3aK

96.

• So in 0.20 M analytical concentration of H3AsO4:

• [H3AsO4] = 0.17 M;

• [H+] = [H2AsO4] = 3 × 10-2 M

• [HAsO42] = 8 × 10-8 M;

• [AsO43] = 2 × 10-15 M

• [OH] = Kw/[H+] = 3 × 10-13M

Polyprotic Acids

• Sulfuric acid (H2SO4): first ionization is strong

• Ka2 = 1.1 x 10-2

• In concentrated solutions, the first ionization produces all the H3O+

• In dilute solutions (less than 0.0010 M), the second dissociation goes to completion

Acid-Base Properties of a Salt Solution

• One of the successes of the Brønsted-Lowry concept of acids and bases was in pointing out that some ions can act as acids or bases.

A 0.1 M solution has a pH of 11.1 and is therefore fairly basic.

Consider a solution of sodium cyanide, NaCN.

)aq(CN)aq(Na)s(NaCN OH 2


• Sodium ion, Na+, is unreactive with water, but the cyanide ion, CN-, reacts to produce HCN and OH-.

• You can also see that OH- ion is a product, so you would expect the solution to have a basic pH. This explains why NaCN solutions are basic.

• The reaction of the CN- ion with water is referred to as the hydrolysis of CN-.

)aq(OH)aq(HCN )l(OH)aq(CN 2


• The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydronium ion.

The CN- ion hydrolyzes to give the conjugate acid and hydroxide.

The hydrolysis reaction for CN- has the form of a base ionization so you write the Kb expression for it.


Predicting Whether a Salt is Acidic, Basic, or Neutral

• How can you predict whether a particular salt will be acidic, basic, or neutral?

The Brønsted-Lowry concept illustrates the inverse relationship in the strengths of conjugate acid-base pairs.

Consequently, the anions of weak acids (poor proton donors) are good proton acceptors.Anions of weak acids therefore, are basic.


• The NH4+ ion hydrolyzes to the conjugate

base (NH3) and hydronium ion.

• This equation has the form of an acid ionization so you write the Ka expression for it.

)aq(OH)aq(NH )l(OH)aq(NH 3324


• These rules apply to normal salts (those in which the anion has no acidic hydrogen)1. A salt of a strong base and a strong

acid.The salt has no hydrolyzable ions and so gives a neutral aqueous solution.

An example is NaCl.


The anion of the salt is the conjugate of the weak acid. It hydrolyzes to give a basic solution.

An example is NaCN.

2. A salt of a strong base and a weak acid.


The cation of the salt is the conjugate of the weak base. It hydrolyzes to give an acidic solution.

An example is NH4Cl.

3. A salt of a weak base and a strong acid.


Both ions hydrolyze. You must compare the Ka of the cation with the Kb of the anion. If the Ka of the cation is larger the solution is acidic. If the Kb of the anion is larger, the solution is basic.

4. A salt of a weak base and a weak acid.


• To predict the acidity or basicity of a salt, you must examine the acidity or basicity of the ions composing the salt.

Consider potassium acetate, KC2H3O2.

The potassium ion is the cation of a strong base (KOH) and does not hydrolyze.

reaction no)l(OH)aq(K 2


Consider potassium acetate, KC2H3O2.The acetate ion, however, is the anion of a weak acid (HC2H3O2) and is basic.

A solution of potassium acetate is predicted to be basic

OHOHCH )l(OH)aq(OHC 2322232

.

The pH of a Salt Solution

• To calculate the pH of a salt solution would require the Ka of the acidic cation or the Kb of the basic anion.

The ionization constants of ions are not listed directly in tables because the values are easily related to their conjugate species.Thus the Kb for CN- is related to the Ka for HCN.


• To see the relationship between Ka and Kb for conjugate acid-base pairs, consider the acid ionization of HCN and the base ionization of CN-.

)aq(CN)aq(OH )l(OH)aq(HCN 32


Ka

Kb

When these two reactions are added you get the ionization of water.

)aq(OH)aq(OH )l(OH2 32 Kw


)aq(CN)aq(OH )l(OH)aq(HCN 32


Ka

Kb

When two reactions are added, their equilibrium constants are multiplied.Therefore,

)aq(OH)aq(OH )l(OH2 32 Kw

wba KKK


• For a solution of a salt in which only one ion hydrolyzes, the calculation of equilibrium composition follows that of weak acids and bases.

The only difference is first obtaining the Ka or Kb for the ion that hydrolyzes.The next example illustrates the reasoning and calculations involved.

A Problem To Consider• What is the pH of a 0.10 M NaCN solution at 25 oC?

The Ka for HCN is 4.9 x 10-10.

Sodium cyanide gives Na+ ions and CN- ions in solution.Only the CN- ion hydrolyzes.


A Problem To Consider

The CN- ion is acting as a base, so first, we must calculate the Kb for CN-.

Now we can proceed with the equilibrium calculation.

510

14

a

wb 100.2

109.4

100.1KK

K



Let x = [OH-] = [HCN], then substitute into the equilibrium expression.

bK]CN[

]OH][HCN[

52

100.2)x10.0(

x



Solving the equation, you find that

3104.1]OH[x

2.11)104.1log(00.14pOH00.14pH 3

Hence,

As expected, the solution has a pH greater than 7.0.

Homework 106• a. KNO2 → K+ + NO2

: NO2 is a weak base. Ignore K+.

• NO2 + H2O ⇌ HNO2 + OH

Initial 0.12 M 0 ~0

• Equil. 0.12 - x x x

• Kb = = =2.5 × 10-11

• Kb = 2.5 × 10-11 = ≈

• x = [OH] = 1.7 × 10-6 M;

• pOH = 5.77; pH = 8.23

a

w

K

K4

14

100.4

100.1

]NO[

]HNO][OH[

2

2

12.0

2x

106• NaOCl → Na+ + OCl: OCl is a weak base. Ignore Na+.

• OCl + H2O ⇌ HOCl + OH Initial 0.45 M 0 ~0

• Equil. 0.45 – x x x

• Kb = = = 2.9 x 10-7

• Kb = 2.9 x 10-7 = ≈

• x = [OH] = 3.6 × 10-4 M;

• pOH = 3.44;

• pH = 10.56

]OCl[

]OH][HOCl[

a

w

K

K8

14

105.3

100.1

2.9 ×

45.0

2x

106• NH4ClO4 → NH4

+ + ClO4: NH4

+ is a weak acid. ClO4

is the conjugate base of a strong acid. ClO4

has no basic (or acidic) properties.

• NH4+ NH⇌ 3 + H+

• Initial 0.40 M 0 ~0

• Equil. 0.40 – x x x

• Ka = = = 5.6 × 10-10

• Ka = ≈

• x = [H+] = 1.5 × 10-5 M; pH = 4.82;

]NH[

]H][NH[

4

3

40.0

2xb

w

K

K5

14

108.1

100.1

AP Chemistry

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