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Homework #6

Chapter 7 Homework Acids and Bases

20. a) 2H2O(l) ⇌ H3O+(aq) + OH-(aq)

𝐾 = [𝐻3𝑂+][𝑂𝐻−] Or

H2O(l) ⇌ H+(aq) + OH-(aq) 𝐾 = [𝐻+][𝑂𝐻−]

b) HCN(aq) + H2O(l) ⇌ H3O+(aq) + CN-(aq)

𝐾 =[𝐻3𝑂+][𝐶𝑁−]

[𝐻𝐶𝑁]

Or HCN(aq) ⇌ H+(aq) + CN-(aq)

𝐾 =[𝐻+][𝐶𝑁−]

[𝐻𝐶𝑁]

c) NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)

𝐾 =[𝑁𝐻4

+][𝑂𝐻−]

[𝑁𝐻3]

For this question you could have picked any weak acid or base or just used a general weak acid (HA) and weak base (B)

24. a) HClO4 strong acid b) HOCl weak acid c) H2SO4 strong acid (1st deprotonation only) d) H2SO3 weak acid 27. The larger the Ka the stronger the acid (table 7.2).

HClO4 > HClO2 > NH4+ > H2O

Strongest acid Weakest acid Note: HClO4 is a strong acid, therefore, will have the highest Ka value.

H2O is the conjugate acid of OH-. OH- is a strong base. Strong bases have the weakest conjugate acids. (H2O(l) + OH-(aq) OH-(aq) + H2O(l))

28. The larger the Ka the stronger the acid (table 7.2), the stronger the acid the weaker the

conjugate base, therefore, the strongest base will have the conjugate acid with the smallest Ka.

NH3 > ClO2- > H2O > ClO4

- Note: HClO4 is a strong acid, therefore, will have the highest Ka value.

Acids are classified as strong acids if they have a Ka greater than 1. H3O+ is one of the weakest strong acids. Therefore, H3O+ (H2O conjugate base) is one of the weakest strong acid and H2O will be slightly more basic than ClO4

-.

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29. The larger the Ka the stronger the acid. a) HCl strong acid and H2O is a weak acid Kw = Ka (1.0×10-14)

b) HNO2 The Ka of HNO2 (4.0×10-4) > Kw = Ka (1.0×10-14)

c) HCN The Ka HCN (6.2×10-10) > Ka of HOC6H5 (1.6×10-10)

30. The higher the Ka the weaker the conjugate base because 𝐾𝑏 =𝐾𝑤

𝐾𝑎

a) Cl- conjugate base of a strong acid, HCl H2O conjugate base of one of the weakest strong acids H3O+ H2O stronger base

b) H2O conjugate base of one of the weakest strong acids H3O+ NO2

- conjugate base of a weak acid NO2

- stronger base c) CN- conjugate base of the weak acid HCN (Ka = 6.2×10-10) OC6H5

- conjugate base of the weak acid HOC6H5 (Ka = 1.6×10-10) OC6H5

- stronger base 31. a ) CH3CO2H acid H2O base CH3CO2

- conjugate base H3O+ conjugate acid Therefore, H2O and CH3CO2

- are competing for protons. b) CH3CO2

- is the stronger base (H3O+ is a strong acid, therefore, forms a weaker conjugate acid than CH3CO2H)

c) CH3CO2- is a weak base because it is the conjugate base of a weak acid.

Conjugate bases of weak acid are also weak, therefore, the do no fully dissociate in solution.

CH3CO2- + H2O(aq) ⇌ CH3CO2H(aq) + OH-(aq)

33. a) weak acid b) strong acid c) weak base d) strong base e) weak base f) weak acid g) weak acid h) strong base i) strong acid (1st deprotonation) 40. The solution is acidic if [H+]>[OH-] , the solution is basic if [H+]<[OH-], and the solution is

neutral if [H+]=[OH-] . The relationship between [H+] and [OH-] is: 𝐾𝑤 = [𝐻+][𝑂𝐻−] = 1.0 × 10−14

a)

(1.0 × 10−7)[𝑂𝐻−] = 1.0 × 10−14

[𝑂𝐻−] = 1.0 × 10−7𝑀

[H+] = [OH-] neutral solution

b)

(8.3 × 10−16)[𝑂𝐻−] = 1.0 × 10−14

[𝑂𝐻−] = 12 𝑀

[H+] < [OH-] basic solution

c) (12)[𝑂𝐻−] = 1.0 × 10−14

[𝑂𝐻−] = 8.3 × 10−16 𝑀

[H+] > [OH-] acidic solution

d)

(5.4 × 10−5)[𝑂𝐻−] = 1.0 × 10−14

[𝑂𝐻−] = 1.9 × 10−10𝑀

[H+] > [OH-] acidic solution

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41. If the pH < 7 the solution is acidic, if the pH > 7 the solution is basic, and if the pH = 7 the solution is neutral. If the pOH < 7 the solution is basic, if the pOH > 7 the solution is acidic, and if the pOH = 7 the solution is neutral.

The relationship between pH and [H+] is: 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] The relationship between [H+] and [OH-] is: 𝐾𝑤 = [𝐻+][𝑂𝐻−] = 1.0 × 10−14

a) Basic

[𝐻+] = 10−𝑝𝐻 = 10−7.40 = 4.0 × 10−8 𝑀 (4.0 × 10−8)[𝑂𝐻−] = 1.0 × 10−14 [𝑂𝐻−] = 2.5 × 10−7𝑀

b) Basic

[𝐻+] = 10−𝑝𝐻 = 10−15.3 = 5 × 10−16 𝑀 (5 × 10−16)[𝑂𝐻−] = 1.0 × 10−14 [𝑂𝐻−] = 20 𝑀

c) Acidic

[𝐻+] = 10−𝑝𝐻 = 10−1.0 = 10 𝑀 (10)[𝑂𝐻−] = 1.0 × 10−14 [𝑂𝐻−] = 1 × 10−15 𝑀

d) Acidic

[𝐻+] = 10−𝑝𝐻 = 10−3.20 = 6.3 × 10−4 𝑀

(6.3 × 10−4)[𝑂𝐻−] = 1.0 × 10−14

[𝑂𝐻−] = 1.6 × 10−11 𝑀

e) Basic

[𝑂𝐻−] = 10−𝑝𝑂𝐻 = 10−5.0 = 1 × 10−5 𝑀

[𝐻+] (1 × 10−5) = 1.0 × 10−14

[𝐻+] = 1 × 10−9 𝑀

f) Acidic

[𝑂𝐻−] = 10−𝑝𝑂𝐻 = 10−9.60 = 2.5 × 10−10 𝑀

[𝐻+] (1.5 × 10−10) = 1.0 × 10−14

[𝐻+] = 4.0 × 10−5 𝑀

47. a) HNO2 is a weak acid The major species in solution are HNO2 The following equilibrium will occur in solution HNO2(aq) ⇌ NO2

-(aq) + H+(aq) Ka = 4.0×10-4 (Table 7.2) Calculate the concentration of H+ ions at equilibrium

HNO2 NO2- H+

Initial 0.250 M 0 0

Change -x +x +x

Equilibrium 0.250-x X X

𝐾𝑎 =[𝑁𝑂2

−][𝐻+]

[𝐻𝑁𝑂2]=

𝑥𝑥

(0.250 − 𝑥)= 4.0 × 10−4

Since Ka is small assume 0.250 –x = 0.250

𝑥𝑥

(0.250)= 4.0 × 10−4

𝑥 = 0.010 𝑀

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Check assumption (5% rule)

𝐴𝑚𝑜𝑢𝑛𝑡 𝐺𝑎𝑖𝑛𝑒𝑑/𝐿𝑜𝑠𝑡

𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛100% =

0.010

0.250100% = 4% Good

Concentration of H+ [𝐻+] = 𝑥 = 0.010 𝑀

Calculate pH 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(0.010) = 2.00

b) HC2H3O2 is a weak acid The major species in solution HC2H3O2 The following equilibrium will occur in solution

HC2H3O2(aq) ⇌ C2H3O2-(aq) + H+(aq) 𝐾𝑎 = 1.8 × 10−5 (Table7.2)

Calculate the concentration of H+ ions at equilibrium

HC2H3O2 C2H3O2- H+

Initial 0.250 M 0 0

Change -x +x +x

Equilibrium 0.250-x x x

𝐾𝑎 =[𝐶2𝐻3𝑂2

−][𝐻+]

[𝐻𝐶2𝐻3𝑂2]=

𝑥𝑥

(0.250 − 𝑥)= 1.8 × 10−5

Since Ka is small assume 0.250-x = 0.250

1.8 × 10−5 =𝑥𝑥

(0.250)

𝑥 = 0.0021 Check assumption (5% rule)

𝐴𝑚𝑜𝑢𝑛𝑡 𝐺𝑎𝑖𝑛𝑒𝑑/𝐿𝑜𝑠𝑡

𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛100% =

0.0021

0.25100% = 0.84% Good

Concentration of H+ [𝐻+] = 𝑥 = 0.0021 𝑀

Calculate pH 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(0.0021) = 2.68

48. a) HOC6H5 is a weak acid therefore

The major species in solution are HOC6H5 The following equilibrium will occur in solution

HOC6H5(aq) ⇌ OC6H5-(aq) + H+(aq) 𝐾𝑎 = 1.6 × 10−10

Calculate the concentrations of H+ at equilirium

HOC6H5 OC6H5- H+

Initial 0.250 M 0 0

Change -x +x +x

Equilibrium 0.250 – x x x

𝐾𝑎 =[𝑂𝐶6𝐻5

−][𝐻+]

[𝐻𝑂𝐶6𝐻5]=

𝑥𝑥

(0.250 − 𝑥)= 1.6 × 10−10

Since Ka is small assume 0.250-x = 0.250

1.6 × 10−10 =𝑥𝑥

(0.250)

𝑥 = 6.3 × 10−6 Check assumption (5% rule)

𝐴𝑚𝑜𝑢𝑛𝑡 𝐺𝑎𝑖𝑛𝑒𝑑/𝐿𝑜𝑠𝑡

𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛100% =

6.3×10−6

0.25100% = 0.0025% Good

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Concentration H+ [𝐻+] = 𝑥 = 6.3 × 10−6 𝑀

Calculate pH 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(6.3 × 10−6) = 5.20

b) HCN is a weak acid The major species in solution are HCN The following equilibrium will occur in solution

HCN(aq) ⇌ CN-(aq) + H+(aq) 𝐾𝑎 = 6.2 × 10−10 (Table 7.2) Calculate the concentration of H+ ion at equilibrium

HCN CN- H+

Initial 0.250 M 0 0

Change -x +x +x

Equilibrium 0.250-x x x

𝐾𝑎 =[𝐶𝑁−][𝐻+]

[𝐻𝐶𝑁]=

𝑥𝑥

(0.250 − 𝑥)= 6.2 × 10−10

Since Ka is small assume 0.250-x = 0.250

6.2 × 10−10 =𝑥𝑥

(0.250)

𝑥 = 1.2 × 10−5 Check assumption (5% rule)

𝐴𝑚𝑜𝑢𝑛𝑡 𝐺𝑎𝑖𝑛𝑒𝑑/𝐿𝑜𝑠𝑡

𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛100% =

1.2×10−5

0.25100% = 0.0048% Good

Concentration H+ [𝐻+] = 𝑥 = 1.2 × 10−5 𝑀

Calculate pH

𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(1.2 × 10−5) = 4.92 49. HF is a weak acid therefore, the reactions that are occurring in solution are:

HF(aq) ⇌ F-(aq) + H+(aq) 𝐾𝑎 = 7.2 × 10−4 (Table 7.2) Calcualte the concentrations of HF, F-, and H+ at equilibrium (you will also need to calculate the consentration of OH- which will be in the soltion from the following equilibrium H2O(l) ⇌ OH-(aq) + H+(aq))

HF F- H+

Initial 0.020 M 0 0

Change -x +x +x

Equilibrium 0.020-x x x

𝐾𝑎 =[𝐹−][𝐻+]

[𝐻𝐹]=

𝑥𝑥

(0.250 − 𝑥)= 7.2 × 10−4

1.44 × 10−5 − 7.2 × 10−4𝑥 = 𝑥2 𝑥2 + 7.2 × 10−4𝑥 − 1.44 × 10−5 = 0

𝑥−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎=

−7.2 × 10−4 ± √(7.2 × 10−4)2 − −5.76 × 10−5

2= 0.0035 𝑎𝑛𝑑 − 0.0042

x is 0.035 M because concentration cannot be negative. Find the final concentrations

[𝐻𝐹] = 0.20 − 𝑥 = 0.020 − 0.0035 = 0.017 𝑀 [𝐹−] = [𝐻+] = 𝑥 = 0.0035 𝑀

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Calculate the concentration of OH- from water equilibrium 𝐾𝑤 = [𝐻+][𝑂𝐻−] = 1.0 × 10−14 (0.0035)[𝑂𝐻−] = 1.0 × 10−14 [𝑂𝐻−] = 2.9 × 10−12 𝑀

Calculate pH 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(0.0035) = 2.46

51. a) HA is a weak acid. There are more HA molecules (green and white atoms

together) than H+ ions (single white atom) and A- ions (single green atoms). This shows that the acid did not completely dissociate making it a weak acid.

b) Calculate the percentage dissociated In the picture there are 9 particles that have note dissociated and 1 particle that has 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑑

𝑡𝑜𝑡𝑎𝑙 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠100% =

1

1 + 9100% = 10%

Calculate the equilibrium constant The reaction of interest is HA(aq) ⇌ H+(aq) + A-(aq)

HA H+ A-

Initial 0.20 M 0 0

Change -x +x +x

Equilibrium 0.20− 𝑥=

9

𝑁𝑎𝑉 𝑥 =

1

𝑁𝑎𝑉 𝑥 =

1

𝑁𝑎𝑉

Note from the picture at equilibrium the concentration of HA is 9

𝑁𝑎

𝑉=

9

𝑁𝑎𝑉 Na is

Avogadro’s number and is used to convert the 9 particles to number of moles of

HA. The concentration of H+ and A- are 1

𝑁𝑎𝑉

Solve for V

0.20 − 𝑥 =9

𝑁𝐴𝑉

0.20 −1

𝑁𝑎𝑉=

9

𝑁𝑎𝑉

𝑉 =10

𝑁𝑎0.20=

50

𝑁𝑎

Solve for K

𝐾 =[𝐻+][𝐴−]

[𝐻𝐴]=

1

𝑁𝑎( 50𝑁𝑎

)

1

𝑁𝑎( 50𝑁𝑎

)

9

𝑁𝑎 (50𝑁𝑎

)

=1

(50)(9)= 0.0022

52. HC2H2ClO2 is a weak acid, therefore, the equilibrium established is:

HC2H2ClO2(aq) ⇌C2H2ClO2-(aq) + H+(aq) 𝐾𝑎 = 1.35 × 10−3

Find the concentration of H+ at equilirbium

HC2H2ClO2 C2H2ClO2- H+

Initial 0.10 M 0 0

Change -x x x

Equilibrium 0.10-x x x

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𝐾𝑎 =[𝐶2𝐻2𝐶𝑙𝑂2

−][𝐻+]

[𝐻𝐶2𝐻2𝐶𝑙𝑂2]=

𝑥𝑥

(0.10 − 𝑥)= 1.35 × 10−3

1.4 × 10−4 − 1.35 × 10−3𝑥 = 𝑥2 𝑥2 + 1.35 × 10−3𝑥 − 1.4 × 10−4 = 0

𝑥−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎=

−1.35 × 10−3 ± √(1.35 × 10−3)2 − −5.6 × 10−4

2= 0.011 𝑎𝑛𝑑 − 0.013

x is 0.011 M because concentration cannot be negative. [𝐻+] = 𝑥 = 0.011 𝑀

Calculate pH 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(0.011) = 1.96

53. HIO3 is a weak acid, therefore, the equilibrium established is:

HIO3(aq) ⇌IO3-(aq) + H+(aq) 𝐾𝑎 = 0.17

Find the concentration of H+ at equilibrium

HIO3 IO3- H+

Initial 0.010 M 0 0

Change -x X x

Equilibrium 0.010-x x x

𝐾𝑎 =[𝐼𝑂3

−][𝐻+]

[𝐻𝐼𝑂3]=

𝑥𝑥

(0.10 − 𝑥)= 0.17

0.0017 − 0.17𝑥 = 𝑥2 𝑥2 + 0.17𝑥 − 0.0017 = 0

𝑥−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎=

−0.17 ± √(0.17)2 − −0.0068

2= 0.0094 𝑎𝑛𝑑 − 0.018

x is 0.0094 M because concentration cannot be negative. [𝐻+] = 𝑥 = 0.0094 𝑀

Calculate pH 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(0.0094) = 2.03

55. 0.56 𝑔 𝐶6𝐻5𝐶𝑂2𝐻 ( 1 𝑚𝑜𝑙 𝐶6𝐻5𝐶𝑂2𝐻

122.13 𝑔 𝐶6𝐻5𝐶𝑂2𝐻) = 0.0046 𝑚𝑜𝑙 𝐶6𝐻5𝐶𝑂2𝐻

𝑀 =𝑛

𝑉=

0.0046 𝑚𝑜𝑙

1.0 𝐿 = 0.0046 𝑀

C6H5CO2H is a weak acid, therefore, the equilibrium established is:

C6H5CO2H(aq) ⇌C6H5CO2-(aq) + H+(aq) 𝐾𝑎 = 6.4 × 10−5

Calcualte the equilirium concnetrations

C6H5CO2H C6H5CO2- H+

Initial 0.0046 M 0 0

Change -x +x +x

Equilibrium 0.0046-x X X

𝐾𝑎 =[𝐶6𝐻5𝐶𝑂2

−][𝐻+]

[𝐻𝐶6𝐻5𝐶𝑂2]=

𝑥𝑥

(0.0046 − 𝑥)= 6.4 × 10−5

2.9 × 10−7 − 6.4 × 10−5𝑥 = 𝑥2 𝑥2 + 6.4 × 10−5𝑥 − 2.9 × 10−7 = 0

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𝑥−𝑏 ± √𝑏2 − 4𝑎𝑐

2𝑎=

−6.4 × 10−5 ± √(6.4 × 10−5)2 − −1.2 × 10−6

2= 5.2 × 10−4 𝑎𝑛𝑑 − 5.8 × 10−4

x is 5.2×10-4 M because concentration cannot be negative. Find the final concentrations

[𝐻𝐶6𝐻5𝐶𝑂2] = 0.0046 − 𝑥 = 0.0046 − 5.2 × 10−4 = 0.0041𝑀 [𝐶6𝐻5𝐶𝑂2

−] = [𝐻+] = 𝑥 = 5.2 × 10−4 𝑀 Calculate the concentration of OH- from water equilibrium

𝐾𝑤 = [𝐻+][𝑂𝐻−] = 1.0 × 10−14 (5.2 × 10−4 𝑀)[𝑂𝐻−] = 1.0 × 10−14 [𝑂𝐻−] = 1.9 × 10−11 𝑀

Calculate pH 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(5.2 × 10−4) = 3.28

56. C6H5CO2H is a solid before it is put into water. Therefore, two equilibriums are

established. C6H5CO2H(s) ⇌ C6H5CO2H(aq)

C6H5CO2H(aq) ⇌ C6H5CO2-(aq) + H+(aq) 𝐾𝑎 = 6.4 × 10−5

They want to know how much C6H5CO2H will dissolve. This will be the amount of C6H5CO2H that starts in the second equilibrium which we will denote as s.

C6H5CO2H C6H5CO2- H+

Initial S 0 0

Change -x +x +x

Equilibrium s-x x x

The pH of the solution will give us the equilibrium concentration of the H+ ions 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] [𝐻+] = 10−𝑝𝐻 = 10−2.80 = 0.0016 𝑀

C6H5CO2H C6H5CO2- H+

Initial s 0 0

Change -0.0016 +0.0016 +0.0016

Equilibrium s-0.0016 0.0016 0.0016

𝐾𝑎 =[𝐶6𝐻5𝐶𝑂2

−][𝐻+]

[𝐶6𝐻5𝐶𝑂2𝐻]=

(0.0016)(0.0016)

(𝑆 − 0.0016)= 6.4 × 10−5

𝑆 = 0.042 𝑀 Convert to grams per 100 ml

𝑆 = (0.042 𝑚𝑜𝑙

1 𝐿) (

1 𝐿

1000 𝑚𝐿) (

122.13 𝑔 𝐶6𝐻5𝐶𝑂2𝐻

1 𝑚𝑜𝑙 𝐶6𝐻5𝐶𝑂2𝐻) = 0.0051

𝑔

𝑚𝐿

Therefore the water solubility per 100 ml is 0.51 g per 100 ml 61. a) The first solution contains 0.10 M HCl (strong acid) and 0.10 M HOCl (weak

acid). Since there is both a strong and weak acid with equal concentrations of each the number of H+ ions from the weak acid will be minimal, therefore, the pH will be completely from the strong acid.

𝑝𝐻 = − log[𝐻+] = − log(0.10) = 1.00 This can be proven using an ICE table HOCl(aq) ⇌ H+(aq) + OCl-(aq) 𝐾𝑎 = 3.5 × 20−8

HOCl OCl- H+

9

Initial 0.10 M 0 0.10

Change -x +x +x

Equilibrium 0.10-x x 0.10+x

𝐾𝑎 =[𝑂𝐶𝑙−][𝐻+]

[𝐻𝑂𝐶𝑙]=

𝑥(0.10 + 𝑥)

(0.10 − 𝑥)= 3.5 × 10−8

Since Ka is small assume 0.10-x = 0.10 and 0.10+x=0.10 𝑥(0.10)

(0.10)= 3.5 × 10−8

𝑥 = 3.5 × 10−8 Check assumption (5% rule)

𝐴𝑚𝑜𝑢𝑛𝑡 𝐺𝑎𝑖𝑛𝑒𝑑/𝐿𝑜𝑠𝑡

𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛100% =

3.5×10−8

0.10100% = 3.5 × 10−5%

Good Concentration H+

[𝐻+] = 0.10 − 𝑥 = 0.10 − 3.5 × 10−8 = 0.10 𝑀 Calculate pH

𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(0.10) = 1.00 b) The second solution contains 0.050 M HNO3 (strong acid) and 0.50 M HC2H3O2

(weak acid). Since there is both a strong and weak acid and even though the concentration of the weak acid is 10x more than the concentration of the strong acid the number of H+ ions from the weak acid will be minimal, therefore, the pH will be completely from the strong acid.

𝑝𝐻 = − log[𝐻+] = − log 0.050 = 1.30 This can be proven using an ICE table

HC2H3O2(aq) ⇌ H+(aq) + C2H3O2-(aq) 𝐾𝑎 = 1.8 × 20−5

HC2H3O2 C2H3O2- H+

Initial 0.50 M 0 0.050

Change -x +x +x

Equilibrium 0.50-x x 0.050+x

𝐾𝑎 =[𝐶2𝐻3𝑂2

−][𝐻+]

[𝐻𝐶2𝐻3𝑂2]=

𝑥(0.050 + 𝑥)

(0.50 − 𝑥)= 1.8 × 10−5

Since Ka is small assume 0.50-x = 0.50 and 0.050+x=0.050 𝑥(0.050)

(0.50)= 1.8 × 10−5

𝑥 = 1.8 × 10−5 Check assumption (5% rule) Note:0.050<0.50 therefore is the assumption works for 0.050-x then it will work form 0.50-x

𝐴𝑚𝑜𝑢𝑛𝑡 𝐺𝑎𝑖𝑛𝑒𝑑/𝐿𝑜𝑠𝑡

𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛100% =

1.8×10−4

0.050100% = 0.36% Good

Concentration H+ [𝐻+] = 0.050 + 𝑥 = 0.050 + 1.8 × 10−4 = 0.050 𝑀

Calculate pH 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(0.50) = 1.30

10

65. HOCN is a weak acid, therefore, the following equilibrium is established: HOCN(aq) ⇌ OCN-(aq) + H+(aq)

HOCN OCN- H+

Initial 1.00×10-2 M 0 0

Change -x +x +x

Equilibrium 1.00×10-2-x x x

The pH of the solution at equilibrium will allow us to calculate the concentration of H+ ions at equilibrium =x

𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] [𝐻+] = 10−𝑝𝐻 = 10−2.77 = 0.0017 𝑀

Updating the ICE table gives

HOCN OCN- H+

Initial 1.00×10-2 M 0 0

Change -0.0017 +0.0017 +0.0017

Equilibrium 0.0083 0.0017 0.0017

𝐾𝑎 =[𝑂𝐶𝑁−][𝐻+]

[𝐻𝑂𝐶𝑁]=

(0.0017)(0.0017)

(0.0083)= 3.5 × 10−4

69. The larger the Kb the stronger the base. Therefore the strongest base will have the

largest Kb. NH3 > C5H5N > H2O > NO3

- Strongest base Weakest base

Note: HNO3 is a strong acid. The stronger the acid the weaker the conjugate base. Therefore, NO3

- is a very weak base. Note: H2O is the conjugate base of (H3O+). H3O+ is one of the weakest strong

acids. Therefore, H2O is only slightly more basic than NO3-

70. The smaller the Kb the stronger the conjugate acid. HNO3 > C5H5NH+ > NH4

+ > H2O Strongest acid Weakest acid Note: HNO3 is a strong acid. Note: C5H5N and NH3 are both weak bases, therefore, will have weak conjugate

acids. The Kb of NH3 > kb of C5H5N resulting in C5H5NH+ being a stronger acid than NH4

+. Note: H2O is the conjugate acid of OH-. OH- is a strong base. Strong bases have

the weakest conjugate acids. (H2O(l) + OH-(aq) OH-(aq) + H2O(l))

74. a) Ca(OH)2 is a strong base (fully dissociates)

[𝑂𝐻−] = 0.00040 𝑀 𝐶𝑎(𝑂𝐻)2 (2 𝑚𝑜𝑙 𝑂𝐻−

1 𝑚𝑜𝑙 𝐶𝑎(𝑂𝐻)2 ) = 0.00080 𝑀 𝑂𝐻−

𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻−] = −𝑙𝑜𝑔(0.00080) = 3.10 𝑝𝐻 + 𝑝𝑂𝐻 = 14.00 𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.00 − 3.10 = 10.90

b) KOH is a strong base (fully dissociates)

25 𝑔 𝐾𝑂𝐻 (1 𝑚𝑜𝑙 𝐾𝑂𝐻

56.11 𝑔 𝐾𝑂𝐻) (

1 𝑚𝑜𝑙 𝑂𝐻−

1 𝑚𝑜𝑙 𝐾𝑂𝐻) = 0.45 𝑚𝑜𝑙 𝑂𝐻−

[𝑂𝐻−] =𝑛

𝑉=

0.45 𝑚𝑜𝑙

1.00 𝐿= 0.45 𝑀

11

𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻−] = −𝑙𝑜𝑔(0.45) = 0.35 𝑝𝐻 + 𝑝𝑂𝐻 = 14.00 𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.00 − 0.35 = 13.65

c) NaOH is a strong base (fully dissociates)

150.0 𝑔 𝑁𝑎𝑂𝐻 (1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻

40.00 𝑔 𝑁𝑎𝑂𝐻) (

1 𝑚𝑜𝑙 𝑂𝐻−

1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻) = 3.750 𝑚𝑜𝑙 𝑂𝐻−

[𝑂𝐻−] =𝑛

𝑉=

3.750 𝑚𝑜𝑙

1.000 𝐿= 3.750 𝑀

𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻−] = −𝑙𝑜𝑔(3.750) = −0.5740 𝑝𝐻 + 𝑝𝑂𝐻 = 14.0000 𝑝𝐻 = 14.0000 − 𝑝𝑂𝐻 = 14.0000 − −0.5740 = 14.5740

75. Ba(OH)2 is a strong base (fully dissociates)

Calculate the concentration of OH- 14.00 = 𝑝𝐻 + 𝑝𝑂𝐻 𝑝𝑂𝐻 = 14.00 − 𝑝𝐻 = 14.00 − 10.50 = 3.50 𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻−] [𝑂𝐻−] = 10−𝑝𝑂𝐻 = 10−3.50 = 3.2 × 10−4

Calculate the concentration of Ba(OH)2

3.2 × 10−4 𝑀 (1 𝑚𝑜𝑙 𝐵𝑎(𝑂𝐻)2

1 𝑚𝑜𝑙 𝑂𝐻− ) = 1.6 × 10−4 𝑀 𝐵𝑎(𝑂𝐻)2

79. The presence of nitrogen most often results in a basic molecule. Nitrogen has an

unpaired electron, therefore, bonds easily to H+ ions. 101. The larger the pH the more basic a solution is

a) HI strong acid HF weak acid

NaF (weak base) NaF(s) Na+(aq) + F-(aq)

Na+ + H2O NaOH + H+ (NaOH is a strong base)

Na+ is neutral

F- + H2O HF + OH-

F- is basic NaI (neutral)

NaI(s) Na+(aq) + I-(aq)

Na+ + H2O NaOH + H+ (NaOH is a strong base)

Na+ is neutral

I- + H2O HI + OH- (HI is a strong acid)

I- is neutral HI < HF < NaI < NaF

Increasing pH b) NH4Br (weak acid)

NH4Br(s) NH4+(aq) + Br-(aq)

NH4+ + H2O NH3 + H3O+

NH4+ is acidic

Br- + H2O HBr + OH- (HBr is a strong acid)

Br- is neutral

12

HBr strong acid KBr (neutral)

KBr(s) K+(aq) + Br-(aq)

K+ + H2O KOH + H+ (KOH is a strong base)

K+ is neutral

Br- + H2O HBr + OH- (HBr is a strong acid)

Br- is neutral NH3 weak base HBr < NH4Br < KBr < NH3 Increasing pH c) C6N5NH3NO3 (weak acid)

C6N5NH3NO3(s) C6N5NH3+(aq) + NO3

-(aq)

C6N5NH3+ + H2O C6N5NH2 + H3O+

C6N5NH3+ is acidic

NO3- + H2O HNO3 + OH- (HNO3 is a strong acid)

NO3- is neutral

NaNO3 (neutral) NaNO3(s) Na+(aq) + NO3

-(aq)

Na+ + H2O NaOH + H+ (NaOH is a strong base)

Na+ is neutral

NO3- + H2O HNO3 + OH- (HNO3 is a strong acid)

NO3- is neutral

NaOH strong base HOC6H5 weak acid (Ka = 1.6×10-10) KOC6H5 (weak base)

KOC6H5(s) K+(aq) + OC6H5-(aq)

K+ + H2O KOH + H+ (KOH is a strong base)

K+ is neutral

OC6H5- + H2O HOC6H5 + OH-

OC6H5- is basic

C6H5NH2 weak base (Kb = 3.8×10-10) HNO3 Strong acid There are 2 weak acid C6N5NH3NO3 (C6H5NH3

+) and HOC6H5 Compare the Ka values to determine the weaker acid HOC6H5 Ka = 1.6×10-10

The Ka value of C6H5NH3+ is not given therefore, it must be calculated

C6H5NH2(aq) Kb = 3.8×10-10

𝐾𝑎 =𝐾𝑤

𝐾𝑏=

1.0 × 10−14

3.8 × 10−10= 2.6 × 10−5

Therefore, C6N5NH2NO3 is a stronger acid than HOC6H5

There are 2 weak bases C6N5NH2 and KOC6H5 (OC6H5-)

Compare the Kb values to determine the weaker base C6N5NH2 Kb = 3.8×10-10

The Kb value of OC6H5- is not given therefore, it must be calculated

HOC6H5(aq) Ka = 1.6×10-10

13

𝐾𝑎 =𝐾𝑤

𝐾𝑏=

1.0 × 10−14

1.6 × 10−10= 6.2 × 10−5

Therefore, KOC6H5 is a stronger base than C6N5NH2 HNO3< C6N5NH3NO3 < HOC6H5 < NaNO3< C6H5NH2 < KOC6H5 < NaOH

103. See chart for problem 85. a) Sr(NO3)2 (salt formed form a strong acid and strong base)

Sr(NO3)2 Sr(NO3)2(s) Sr2+(aq) + 2NO3

-(aq)

Sr2+ + 2H2O Sr(OH)2 + 2H+ (Sr(OH)2 is a strong base)

Sr2+ is neutral

NO3- + H2O HNO3 + OH- (HNO3 is a strong acid)

NO3- is neutral

Therefore, Sr(NO3)2 is neutral b) C2H5NH3CN

C2H5NH3CN(s) C2H5NH3+(aq) + CN-(aq)

C2H5NH3+ + H2O C2H5NH2 + H3O+

C2H5NH3+ is acidic

CN- + H2O HCN + OH-

CN- is basic Therefore, need to compare Ka and Kb values of reactions

We are not given the Ka of C2H5NH3+, therefore, we need to calculate it

C2H5NH2(aq) Kb = 5.6×10-4

𝐾𝑎 =𝐾𝑤

𝐾𝑏=

1.0 × 10−14

5.6 × 10−4= 1.8 × 10−11

We are not given the Kb of CN-, therefore, we need to calculate it HCN(aq) Ka = 6.2×10-10

𝐾𝑏 =𝐾𝑤

𝐾𝑎=

1.0 × 10−14

6.2 × 10−10= 1.6 × 10−5

Since Ka < Kb C2H5NH3CN is basic c) C5H5NHF

C5H5NHF(s) C5H5NH+(aq) + F-(aq)

C5H5NH+ + H2O C5H5N + H3O+

C5H5NH+ is acidic

F- + H2O HF + OH-

F- is basic Therefore, need to compare Ka and Kb values of reactions We are not given the Ka of C5H5NH+, therefore, we need to calculate it

C2H5N(aq) Kb = 1.7×10-9

𝐾𝑎 =𝐾𝑤

𝐾𝑏=

1.0 × 10−14

1.7 × 10−9= 5.9 × 10−6

We are not given the Kb of F-, therefore, we need to calculate it HF(aq) Ka = 7.2×10-4

𝐾𝑏 =𝐾𝑤

𝐾𝑎=

1.0 × 10−14

7.2 × 10−4= 1.4 × 10−11

Since Ka > Kb C5H5NHF is acidic

14

d) NH4C2H3O2

NH4C2H3O2(s) NH4+(aq) + C2H3O2

-(aq)

NH4+ + H2O NH3 + H3O+

NH4+ is acidic

C2H3O2- + H2O HC2H3O2 + OH-

C2H3O2- is basic

Therefore, need to compare Ka and Kb values of reactions NH4

+ Ka = 5.6×10-10

We are not given the Kb of C2H3O2- therefore we need to calculate it

HC2H3O2(aq) Ka = 1.8×10-5

𝐾𝑏 =𝐾𝑤

𝐾𝑎=

1.0 × 10−14

1.8 × 10−5= 5.6 × 10−10

Ka = Kb NH4C2H3O2 is neutral e) NaHCO3

NaHCO3(s) Na+(aq) + HCO3-(aq)

Na+ + H2O Na(OH) + H+ (NaOH is a strong base)

Na+ is neutral

HCO3- + H2O H2CO3 + OH-

HCO3- + H2O CO3

2- + H3O+

Since H2CO3 is a polyprotic acid must look at Ka and Kb of the two possible reactions.

HCO3- Ka = 4.8×10-11

We are not given the Kb of HCO3- therefore we need to calculate it

H2CO3(aq) Ka = 4.3×10-7

𝐾𝑏 =𝐾𝑤

𝐾𝑎=

1.0 × 10−14

4.3 × 10−7= 2.3 × 10−8

Since Ka < Kb HCO3- is basic

Therefore, NaHCO3 is basic 105. a) KCl KNO2 KCl(s) K+(aq) + Cl-(aq)

K+ + H2O KOH + H+ (KOH is a strong base)

K+ is neutral

Cl- + H2O HCl + OH- (HCl is a strong acid)

Cl- is neutral Therefore, KCl is neutral

Therefore, the solution will be neutral pH=7.00. For neutral solutions the concentration of H+ and OH- equals 1.0×10-7 M.

b) KF KF(s) K+(aq) + F-(aq)

K+ + H2O KOH + H+ (KOH is a strong base)

K+ is neutral

F- + H2O HF + OH-

F- is basic Therefore, KF is basic

15

Interested in the following reaction F-(aq) + H2O(l) ⇌ HF(aq) + OH-(aq) Problem we do not know Kb HF(aq) 𝐾𝑎 = 7.2 × 10−4

𝐾𝑏 =𝐾𝑤

𝐾𝑎=

1.0×10−14

7.2×10−4 = 1.4 × 10−11

Calculate the concentration of OH- ions

F- HF OH-

Initial 1.0 M 0 0

Change -x +x +x

Equilibrium 1.0-x x x

𝐾𝑏 =[𝑂𝐻−][𝐻𝐹]

[𝐹−]=

𝑥𝑥

1.0 − 𝑥= 1.4 × 10−11

Since Kb is small assume that 1.0-x = 1.0 𝑥𝑥

1.0= 1.4 × 10−11

𝑥 = 3.7 × 10−6 Check assumption (5% rule)

𝐴𝑚𝑜𝑢𝑛𝑡 𝐺𝑎𝑖𝑛𝑒𝑑/𝐿𝑜𝑠𝑡

𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛100% =

3.7×10−6

1.0100% = 3.7 × 10−4% Good

Calculate OH- concentration [𝑂𝐻−] = 𝑥 = 3.7 × 10−6 𝑀

Calculate H+ concentration

[𝐻+][𝑂𝐻−] = 1.0 × 10−14

[𝐻+] =1.0 × 10−14

3.7 × 10−6= 2.7 × 10−9

Calculate pH 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(2.7 × 10−9) = 8.57

109. a) KNO2 KNO2(s) K+(aq) + NO2

-(aq)

K+ + H2O KOH + H+ (KOH is a strong base)

K+ is neutral

NO2- + H2O HNO2 + OH-

NO2- is basic

Therefore, KNO2 is basic Interested in the following reaction

NO2-(aq) + H2O(l) ⇌ HNO2(aq) + OH-(aq)

Problem we don’t know Kb Know

HNO2(aq) 𝐾𝑎 = 4.0 × 10−4

𝐾𝑏 =𝐾𝑤

𝐾𝑎=

1.0×10−14

4.0×10−4 = 2.5 × 10−11

Calculate the concentration of OH- ions

NO2- HNO2 OH-

Initial 0.12 M 0 0

Change -x +x +x

Equilibrium 0.12-x x x

16

𝐾𝑏 =[𝐻𝑁𝑂2][𝑂𝐻−]

[𝑁𝑂2−]

=𝑥𝑥

0.12 − 𝑥= 2.5 × 10−11

Since Kb is small assume that 0.12-x = 0.12 𝑥𝑥

0.12= 2.5 × 10−11

𝑥 = 1.7 × 10−6 𝑀 Check assumption (5% rule)

𝐴𝑚𝑜𝑢𝑛𝑡 𝐺𝑎𝑖𝑛𝑒𝑑/𝐿𝑜𝑠𝑡

𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛100% =

1.7×10−6

0.12100% = 0.0014% Good

Calculate OH+ concentration [𝑂𝐻−] = 𝑥 = 1.7 × 10−6 𝑀

Calculate pH 𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻−] = −𝑙𝑜𝑔(1.7 × 10−6) = 10.56 𝑝𝐻 + 𝑝𝑂𝐻 = 14.00 𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.00 − 5.77 = 8.23

b) NaOCl NaOCl(s) Na+(aq) + OCl-(aq)

Na+ + H2O NaOH + H+ (NaOH is a strong base)

Na+ is neutral

OCl- + H2O HOCl + OH-

OCl- is basic Therefore, NaOCl is basic Interested in the following reaction OCl-(aq) + H2O(l) ⇌ HOCl(aq) + OH-(aq) Problem we don’t know Kb

HOCl(aq) 𝐾𝑎 = 3.5 × 10−8

𝐾𝑏 =1.0 × 10−14

3.5 × 10−8= 2.9 × 10−7

Calculate the concentration of OH- ions

OCl- HOCl OH-

Initial 0.45 M 0 0

Change -x +x +x

Equilibrium 0.45-x X x

𝐾𝑏 =[𝐻𝑂𝐶𝑙][𝑂𝐻−]

[𝑂𝐶𝑙−]=

𝑥𝑥

0.45 − 𝑥= 2.9 × 10−7

Since Kb is small assume that 0.45-x = 0.45 𝑥𝑥

0.45= 2.9 × 10−7

𝑥 = 3.6 × 10−4 𝑀 Check assumption (5% rule)

𝐴𝑚𝑜𝑢𝑛𝑡 𝐺𝑎𝑖𝑛𝑒𝑑/𝐿𝑜𝑠𝑡

𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛100% =

3.6×10−4

0.45100% = 0.080% Good

Calculate OH- concentration [𝑂𝐻−] = 𝑥 = 3.6 × 10−4 𝑀

Calculate pH 𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻−] = −𝑙𝑜𝑔(3.6 × 10−4) = 3.44 𝑝𝐻 + 𝑝𝑂𝐻 = 14.00 𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.00 − 3.44 = 10.56

17

c) NH4OCl4 (a salt from a weak base and a strong acid) NH4OCl4(s) NH4

+(aq) + OCl4-(aq)

NH4+ + H2O NH3 + H3O+

NH4+ is acidic

OCl4- + H2O HOCl4 + OH- (HOCl4 strong acid)

OCl4- is neutral Therefore, NH4OCl4 is acidic

Interested in the following reaction NH4

+(aq) ⇌ NH3(aq) + H+(aq) 𝐾𝑎 = 5.6 × 10−10 Calculate the concentration of H+ ions

NH4+ NH3 H+

Initial 0.40 M 0 0

Change -x +x +x

Equilibrium 0.40-x x X

𝐾𝑎 =[𝑁𝐻3][𝐻+]

[𝑁𝐻4+]

=𝑥𝑥

0.40 − 𝑥= 5.6 × 10−10

Since Kb is small assume that 0.40-x = 0.40 𝑥𝑥

0.40= 5.6 × 10−10

𝑥 = 1.5 × 10−5 𝑀 Check assumption (5% rule)

𝐴𝑚𝑜𝑢𝑛𝑡 𝐺𝑎𝑖𝑛𝑒𝑑/𝐿𝑜𝑠𝑡

𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛100% =

1.5×10−5

0.40100% = 0.0038% Good

Calculate H+ concentration [𝐻+] = 𝑥 = 1.5 × 10−5

Calculate pH

𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(1.5 × 10−5) = 4.82 131. Because the solution has a pH of 4.6 we know that the solution is acidic, therefore,

NaOH, NH3, and NaCN cannot be the solute.

Because the bulb is bright it means that there are a lot of ions in solution. Therefore, a strong electrolyte is needed. Only HCl, NaOH, NH4Cl, NaCN are strong electrolytes.

NaOH and NaCN were already eliminated because they were basic.

HCl could light the bulb bright because it is a strong acid but the pH of a 1.0 M solution of HCl is

𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(1.0) = 0 Confirm the solution is made from NH4Cl by compare actual to calculated pH values Determine the Ka of the weak acid.

HA(aq) ⇌ A-(aq) + H+(aq) Ka = unknown

HA A- H+

Initial 1.0 M 0 0

Change -x +x +x

Equilibrium 1.0-x x x

18

𝐾𝑎 =[𝐴−][𝐻+]

[𝐻𝐴]=

𝑥𝑥

(1.0 − 𝑥)

Determine x 𝑥 = [𝐻+] 𝑝𝐻 = − log[𝐻+]

4.6 = − log[𝐻+] [𝐻+] = 2.5 × 10−5

𝐾𝑎 =[𝐴−][𝐻+]

[𝐻𝐴]=

(2.5 × 10−5)(2.5 × 10−5)

1.0 − 2.5 × 10−5= 6.3 × 10−10

This Ka of NH4+ is 5.6×10-10 which is close to the calculated value. Therefore the solution

must contain NH4Cl