homework #3 chapter 8 · 1 homework #3 chapter 8 applications of aqueous equilibrium 15. buffer...

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Homework #3

Chapter 8 Applications of Aqueous Equilibrium

15. Buffer solution: A solution that resists change in pH when a small amount of acid

or base is added.

Buffer solutions contain a weak acid and its conjugate base or a weak base and its conjugate acid. For weak acid/conjugate base systems, when a base is added to the solution it reacts with the weak acid to form more conjugate base. When an acid is added to the same solution it reacts with conjugate base to form more weak acid. In both cases the pH remains approximately constant since the number of OH-/H+ ions in solution is approximately constant. A similar system is set up for weak base/ conjugate acid buffers. In buffer systems the larger the amount of weak acid/conjugate base or weak base/conjugate acid the better the buffer. In addition, the best buffers have equal amounts of weak acid and conjugate base or weak base and conjugate acid.

If the buffer solution is made of NaHCO3 and Na2CO3 then the following equation will happen in solution

HCO3-(aq) + OH-(aq) ⇌ CO3

2-(aq) + H2O(l) CO3

2-(aq) + H+(aq) ⇌ HCO3-(aq)

16. Buffer Capacity: An indication of the amount of acid or base that can be added before a

buffer loses its ability to resist the change in pH.

The buffer capacity is greatest when there are equal amount of weak acid and conjugate base or weak base and conjugate acid. In addition, as the overall amount of weak acid and conjugate base goes up the buffering capacity of the solutions go up (same as for weak base and conjugate acid). Therefore, solution C which has 1.0 M solutions of a weak acid and its conjugate base will have the greatest buffering capacity.

The Henderson-Hasselbalch equation allows us to calculate the pH of a buffered system.

𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔 ([𝐴−]

[𝐻𝐴])

Systems with the greatest buffer capacity will have [A-]=[HA] causing the pH=pKa. Therefore, when choosing a buffer system you should choose a system that has a pKa close to what the pH of the overall solution should be. Since all of the 3 systems have equal amounts of [A-] and [HA] the pH for all the systems is the same.

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21. a) Major Species: HC3H5O2 No reaction goes to completion Not a buffer (no conjugate base present)

Use an ICE table to determine pH. HC3H5O2(aq) ⇌ H+(aq) + C3H5O2

-(aq) 𝐾𝑎 = 1.3 × 10−5

HC3H5O2 H+ C3H5O2-

Initial (M) 0.100 0 0

Change (M) -x +x +x

Equilibrium (M) 0.100-x x x

𝐾𝑎 =[𝐻+][𝐶3𝐻5𝑂2

+]

[𝐻𝐶3𝐻5𝑂2]=

𝑥𝑥

(0.100 − 𝑥)= 1.3 × 10−5

Since Ka is small assume 0.100 –x = 0.100 𝑥2

0.100= 1.3 × 10−5

𝑥 = 0.0011 Check assumption

0.0011

0.100100% = 1.1% Good

Concentration of H+ [𝐻+] = 𝑥 = 0.011 𝑀

Calculate the pH 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(0.0011) = 2.96

b) Major Species: Na+ and C3H5O2-

No reaction goes to completion Not a buffer (no conjugate acid present)

Use an ICE table to determine pH. Need to determine Kb of:

C3H5O2-(aq) + H2O(l) ⇌ HC3H5O2(aq) + OH-(aq) Kb =?

𝐾𝑏 =𝐾𝑤

𝐾𝑎=

1.0 × 10−14

1.3 × 10−5= 7.7 × 10−10

C3H5O2- HC3H5O2 OH-

Initial (M) 0.100 0 0

Change (M) -x +x +x

Equilibrium (M) 0.100-x x X

𝐾𝑏 =[𝐻𝐶3𝐻5𝑂2][𝑂𝐻−]

[𝐶3𝐻5𝑂2−]

=𝑥𝑥

(0.100 − 𝑥)= 7.7 × 10−10

Since Kb is small assume 0.100 –x = 0.100 𝑥2

0.100= 7.7 × 10−10

𝑥 = 8.8 × 10−6 Check assumption

8.8×10−6

0.100100% = 0.0088% Good

Concentration of OH- [𝑂𝐻−] = 𝑥 = 8.8 × 10−6 𝑀

Calculate pOH 𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻−] = −𝑙𝑜𝑔(8.8 × 10−6) = 5.06

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Calculate pH 𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.00 − 5.06 = 8.94

c) For pure water [H+] = [OH-] = 1.0×10-7 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(1.0 × 10−7) = 7.00

d) Major Species: HC3H5O2, Na+, and C3H5O2

- No reaction goes to completion Buffer

Use Henderson-Hasselbalch equation to solve for pH.


[𝐻𝐴]) = −𝑙𝑜𝑔(1.3 × 10−5) + 𝑙𝑜𝑔 (

0.100

0.100) = 4.89

You could have also used and ice table to solve for pH.

HC3H5O2(aq) ⇌ H+(aq) + C3H5O2-(aq) 𝐾𝑎 = 1.3 × 10−5

HC3H5O2 H+ C3H5O2-

Initial (M) 0.100 0 0.100

Change (M) -x +x +x

Equilibrium (M) 0.100-x x 0.100+x


+]


𝑥(0.100 + 𝑥)

(0.100 − 𝑥)= 1.3 × 10−5

Since Ka is small assume 0.100 –x = 0.100 and 0.100+x=0.100 𝑥(0.100)

(0.100)= 1.3 × 10−5


1.3×10−5

0.100100% = 0.013% Good

Concentration H+ [𝐻+] = 𝑥 = 1.3 × 10−5 𝑀

Calculate pH 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(1.3 × 10−5) = 4.89

22. a) Major Species: H+

, Cl-, and HC3H5O2 No reaction goes to completion Not a buffer (no conjugate acid present)

The majority of the H+ ions are coming from the strong acid. Therefore, calculate pH as you would for a strong acid.

𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(0.0020) = 1.70 We can check this by using the long way


HC3H5O2 H+ C3H5O2-

Initial (M) 0.100 0.020 0

Change (M) -x +x +x

Equilibrium (M) 0.100-x 0.020+x X


+]


𝑥(0.020 + 𝑥)

(0.100 − 𝑥)= 1.3 × 10−5

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Since Ka is small assume 0.020+x = 0.020 and 0.100-x = 0.100 𝑥(0.020)

(0.100)= 1.3 × 10−5

𝑥 = 6.5 × 10−5 Check assumptions

6.5×10−5

0.020100% = 0.33% Good

6.5×10−5

0.100100% = 0.065% Good

Concentration of H+ [𝐻+] = 0.020 + 𝑥 = 0.020 + 6.5 × 10−5 = 0.020 𝑀

Calculate the pH 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(0.020) = 1.70

b) Major Species: H+, Cl-, and C3H5O2

-V Yes a reaction goes to completion H+ + C3H5O2

- HC3H4O2(aq)

C3H5O2- H+ HC3H5O2

Initial (mol) 0.100 0.020 0

Final (mol) 0.080 0 0.020

Major Species after reaction: C3H5O2- and HC3H5O2

Buffer


[𝐻𝐴]) = −𝑙𝑜𝑔(1.3 × 10−5) + 𝑙𝑜𝑔 (

0.080 𝑚𝑜𝑙𝑉


)

= 5.49 c) Major Species: H+ and Cl- No reaction goes to completion Not a buffer

𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(0.020) = 1.70 d) Major Species: H+, Cl-, Na+, and C3H5O2

-

Yes a reaction goes to completion C3H5O2

-(aq) + H+(aq) HC3H5O2(aq)

C3H5O2- H+ HC3H5O2

Initial (mol) 0.100 0.020 0.100

Final (mol) 0.080 0 0.120

Major Species: Cl-, Na+, C3H5O2- and HC3H5O2

Buffer


[𝐻𝐴]) = −𝑙𝑜𝑔(1.3 × 10−5) + 𝑙𝑜𝑔 (



)

= 4.71

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23. a) Major Species: HC3H5O2, Na+, and OH- Yes a reaction goes to completion

HC3H5O2(aq) + OH-(aq) C3H5O2-(aq) + H2O(l)

HC3H5O2 OH- C3H5O2-

Initial (mol) 0.100 0.020 0

Final (mol) 0.080 0 0.020

Major Species: HC3H5O2, Na+, and C3H5O2-

Buffer


[𝐻𝐴]) = −𝑙𝑜𝑔(1.3 × 10−5) + 𝑙𝑜𝑔 (



)

= 4.28 b) Major Species: C3H5O2

-, Na+, and OH- No reaction goes to completion. Not a buffer. The majority of the OH- ions are from the strong base. Therefore, use the strong base concentration to calculate pH

𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻−] = −𝑙𝑜𝑔(0.020) = 1.70 𝑝𝐻 + 𝑝𝑂𝐻 = 14.00 𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.00 − 1.70 = 12.30

We can check this C3H5O2

-(aq) + H2O(l)⇌ OH-(aq) + HC3H5O2(aq) 𝐾𝑏 = 7.7 × 10−10

C3H5O2- OH- HC3H5O2

Initial (M) 0.100 0.020 0

Change (M) -x +x +x

Equilibrium (M) 0.100-x 0.020+x X

𝐾𝑏 =[𝑂𝐻−][𝐻𝐶3𝐻5𝑂2]

[𝐶3𝐻5𝑂2−]

=𝑥(0.020 + 𝑥)

(0.100 − 𝑥)= 7.7 × 10−10

Since K is small assume 0.020+x = 0.020 and 0.100-x = 0.100 𝑥(0.020)

(0.100)= 7.7 × 10−10

𝑥 = 3.9 × 10−9 Check assumptions

3.9×10−9

0.020100% = 2.0 × 10−5% Good

3.9×10−9

0.100100% = 3.9 × 10−6% Good

Concentration of OH-

[𝑂𝐻−] = 0.20 + 𝑥 = 0.020 + 3.9 × 10−9 = 0.020 𝑀 Calculate the pH

𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻−] = −𝑙𝑜𝑔(0.020) = 1.70 𝑝𝐻 + 𝑝𝑂𝐻 = 14.00 𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.00 − 1.70 = 12.30 c) Major Species: Na+, and OH-

No reaction goes to completion Not a Buffer

𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻−] = −𝑙𝑜𝑔(0.020) = 1.70

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𝑝𝐻 + 𝑝𝑂𝐻 = 14.00 𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14 − 1.70 = 12.30 d) Major Species: HC3H5O2, C3H5O2

-, Na+, and OH-

Yes a reaction goes to completion HC3H5O2(aq) + OH-(aq) C3H5O2

-(aq) + H2O(l)

HC3H5O2 OH- C3H5O2-

Initial (mol) 0.100 0.020 0.100

Final (mol) 0.080 0 0.120

Major Species: HC3H5O2, C3H5O2-, and Na+

Buffer


[𝐻𝐴]) = −𝑙𝑜𝑔(1.3 × 10−5) + 𝑙𝑜𝑔 (



)

= 5.06 27. Major Species: HF, K+, and F- No reaction goes to completion Buffer Ka = 7.2×10-4 (Table 7.2)


[𝐻𝐴]) = −𝑙𝑜𝑔(7.2 × 10−4) + +𝑙𝑜𝑔 (

1.00

0.60) = 3.36

29. Major Species: HF, K+, F-, Na+, and OH- Yes a reaction goes to completion

HF(aq) + OH-(aq) F-(aq) + H2O(l)

HF OH- F-

Initial (mol) 0.60 0.10 1.00

Final (mol) 0.50 0 1.10

Major Species: HF, K+, F-, and Na+ Buffer


[𝐻𝐴]) = −𝑙𝑜𝑔(7.2 × 10−4) + 𝑙𝑜𝑔 (



) = 3.49

Major Species: HF, K+, F-, H+, and Cl- Yes a reaction goes to completion

F-(aq) + H+(aq) HF(aq)

F- H+ HF

Initial (mol) 1.00 0.20 0.60

Final (mol) 0.80 0 0.80

Major Species: HF, K+, F-, and Na+ Buffer


[𝐻𝐴]) = −𝑙𝑜𝑔(7.2 × 10−4) + 𝑙𝑜𝑔 (



) = 3.14

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35. All of the solutions are buffer solutions because there is a weak base (C5H5N) and its conjugate acid (C5H5NH+) present. Therefore, you can use Henderson-Hasselbalch equation.


[𝐻𝐴])

[𝐴−]

[𝐻𝐴]= 10𝑝𝐻−𝑝𝐾𝑎

Need to determine pKa of C5H5NH+

𝐾𝑎 =𝐾𝑤

𝐾𝑏=

1.0×10−14

1.7×10−9 = 5.9 × 10−6

Calculate pKa

𝑝𝐾𝑎 = −𝑙𝑜𝑔(𝐾𝑎) = −𝑙𝑜𝑔(5.9 × 10−6) = 5.23

a) [𝐴−]

[𝐻𝐴]= 10𝑝𝐻−𝑝𝐾𝑎 = 104.50−5.23 = 0.19

b) [𝐴−]

[𝐻𝐴]= 10𝑝𝐻−𝑝𝐾𝑎 = 105.00−5.23 = 0.59

c) [𝐴−]

[𝐻𝐴]= 10𝑝𝐻−𝑝𝐾𝑎 = 105.23−5.23 = 1.0

d) [𝐴−]

[𝐻𝐴]= 10𝑝𝐻−𝑝𝐾𝑎 = 105.50−5.23 = 1.9

37. Major Species: Na+, C2H3O2

-, H+, and Cl- Yes a reaction will go to completion H+ + C2H3O2

- HC2H3O2

H+ C2H3O2- HC3H5O2

Initial (mol) x 1.0 0

Final (mol) 0 1.0-x x

Major Species: C2H3O2- , HC2H3O2, Na+, and Cl-

Buffer a) When the pH = pKa [HC2H3O2] = [C2H3O2

-]

[𝐴−]

[𝐻𝐴]= 1 =

1.0 𝑚𝑜𝑙 − 𝑥𝑉𝑥𝑉

𝑥 = 0.5 𝑚𝑜𝑙 0.5 moles of HCl is needed

b) Ka = 1.8×10-5

Still have a buffer


[𝐻𝐴])

𝑝𝐾𝑎 = −𝑙𝑜𝑔(1.8 × 10−5) = 4.74

4.20 = 4.74 + 𝑙𝑜𝑔 (


)

0.29 =1.0 𝑚𝑜𝑙 − 𝑥

𝑥

𝑥 = 0.78 𝑀 0.78 moles of HCl is needed.

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c) Still have a buffer


[𝐻𝐴])

5.00 = 4.74 + 𝑙𝑜𝑔 (


)

1.8 =1.0 𝑚𝑜𝑙 − 𝑥

𝑥

𝑥 = 0.36 𝑀 0.36 moles of HCl is needed.

39. Major Species: Na=, C2H3O2

-, and HC2H3O2 No reaction goes to completion Buffer

𝐾𝑎 = 1.8 × 10−5


[𝐻𝐴])

5.00 = −𝑙𝑜𝑔(1.8 × 10−5) + 𝑙𝑜𝑔 ([𝐴−]

0.200 𝑀)

[𝐴−] = 0.36 𝑀 They asked for the mass not the molarity

0.500 𝐿 𝐶2𝐻3𝑂2− (

0.36 𝑚𝑜𝑙 𝐶2𝐻3𝑂2−

1 𝐿 𝐶2𝐻3𝑂2−

) (1 𝑚𝑜𝑙 𝑁𝑎𝐶2𝐻3𝑂2

1 𝑚𝑜𝑙 𝐶2𝐻3𝑂2− ) (

82.04 𝑔 𝑁𝑎𝐶2𝐻3𝑂2

1 𝑚𝑜𝑙 𝑁𝑎𝐶2𝐻3𝑂2)

= 15 𝑔 𝑁𝑎𝐶2𝐻3𝑂2

40. a) Major Species: H+, Cl-, NH3, and NH4+

A reaction will go to completion NH3(aq) + H+(aq) NH4

+(aq)

NH3 H+ NH4+

Initial (mol) 0.0125 0.010 0.0375

Final (mol) 0.0025 0 0.0475

Major Species: Cl-, NH3, and NH4+

Buffer


[𝐻𝐴]) = −𝑙𝑜𝑔(5.6 × 10−10) + 𝑙𝑜𝑔 (

0.0025 𝑚𝑜𝑙𝑉

0.0475 𝑚𝑜𝑙𝑉

)

= 7.97 b) Major Species: H+, Cl-, NH3, and NH4

+ A reaction will go to completion

NH3(aq) + H+(aq) NH4+(aq)

NH3 H+ NH4+

Initial (mol) 0.125 0.010 0.375

Final (mol) 0.115 0 0.385

Major Species: Cl-, NH3, and NH4+

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Buffer


[𝐻𝐴]) = −𝑙𝑜𝑔(5.6 × 10−10) + 𝑙𝑜𝑔 (



)

= 8.73 The pH of the original buffer solutions are:

a) 𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔 ([𝐴−]

[𝐻𝐴]) = −𝑙𝑜𝑔(5.6 × 10−10) + 𝑙𝑜𝑔 (

0.050 𝑀

0.15 𝑀) = 8.77

b) 𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔 ([𝐴−]

[𝐻𝐴]) = −𝑙𝑜𝑔(5.6 × 10−10) + 𝑙𝑜𝑔 (

0.50 𝑀

1.50 𝑀) = 8.77

Therefore, they start out with the same pH. Solution b is able to keep the pH closer to the original pH because it has a greater buffer capacity due to the larger amounts of NH3 and NH4Cl.

42. Major Species: HNO3, Na+, and NO2

No reaction goes to completion Buffer


[𝐻𝐴])

𝑝𝐾𝑎 = −𝑙𝑜𝑔(𝐾𝑎) = −𝑙𝑜𝑔(4.0 × 10−4) = 3.40 (Table 7.2) Total volume x + y = 1.00 x = volume of HNO2

y = volume of NO2-

Molarity of HNO2 in final solution

𝑀 =𝑛

𝑉=

(0.50 𝑀)𝑥

𝑥 + 𝑦=

(0.50 𝑀)𝑥

1.0 𝐿

Molarity of NO2- in final solution

𝑀 =𝑛

𝑉=

(0.50 𝑀)𝑦

𝑥 + 𝑦=

(0.50 𝑀)𝑦

1.0 𝐿

Use the Henderson-Hasselbalch equation to find another relationship between x and y


[𝐻𝐴])

3.55 = 3.40 + 𝑙𝑜𝑔 (

(0.50 𝑀)𝑦1.0 𝐿

(0.50 𝑀)𝑥1.0 𝐿

)

3.55 = 3.40 + 𝑙𝑜𝑔 (𝑦

𝑥)

𝑦 = 1.4𝑥 The total volume of the solution is 1 L (x+y=1). Use this equation to solve for x and y.

𝑦 = 1.4𝑥 𝑥 + 𝑦 = 1.00 𝑥 + 1.4𝑥 = 2.4𝑥 = 1.00 𝑥 = 0.42 𝐿 = 𝑉𝐻𝑁𝑂2

𝑦 = 0.58 𝐿 = 𝑉𝑁𝑂2−

10

43. a) Major Species: H2PO4- and HPO4

2- No reaction goes to completion. Buffer


[𝐻𝐴])

𝑝𝐾𝑎 = −𝑙𝑜𝑔(𝐾𝑎) = −𝑙𝑜𝑔(6.2 × 10−8) = 7.21

7.15 = 7.21 + 𝑙𝑜𝑔 ([𝐻𝑃𝑂4

2−]

[𝐻2𝑃𝑂4−]

)

[𝐻𝑃𝑂42−]

[𝐻2𝑃𝑂4−]

= 0.87

[𝐻2𝑃𝑂4−]

[𝐻𝑃𝑂42−]

=1

0.87= 1.1

b) The pH of intracellular fluid is ~7.15 (part b). The pKa of H3PO4 is 2.12. The best buffers have equal amounts of A- and HA, therefore, the pH = pKa. In order to have a pH of 7.15 with a solution containing H3PO4/H2PO4

- there would have to be much more H2PO4

- in solution that H3PO4 making it an ineffective buffer. 44. a) Major Species: H2CO3 and HCO3

- No reaction goes to completion Buffer


[𝐻𝐴])

𝑝𝐾𝑎 = −𝑙𝑜𝑔(𝐾𝑎) = −𝑙𝑜𝑔(4.3 × 10−7) = 6.37

7.40 = 6.37 + 𝑙𝑜𝑔 ([𝐻𝐶𝑂3

−]

[𝐻2𝐶𝑂3]) = 6.37 + 𝑙𝑜𝑔 (

[𝐻𝐶𝑂3−]

0.0012 𝑀)

[𝐻𝐶𝑂3−] = 0.013 𝑀

46. a) Major Species: K+, OH-, CH3NH3

+, and Cl- Yes a reaction goes to completion OH-(aq) + CH3NH3

+(aq) H2O(l) + CH3NH2(aq)

OH- CH3NH3+ CH3NH2

Initial (mol) 0.1 0.1 0

Final (mol) 0 0 0.1

Major Species: K+, CH3NH2, and Cl- Not a buffer (weak base) b) Major Species: K+, OH-, and CH3NH2

No reaction goes to completion Not a buffer (weak base and strong base)

c) Major Species: K+, OH-, CH3NH3+, and Cl-

Yes a reaction goes to completion OH-(aq) + CH3NH3


OH- CH3NH3+ CH3NH2


Final (mol) 0.1 0 0.1

Major Species: K+, OH-, CH3NH2, and Cl- Not a buffer (weak base and strong base)

11

d) Major Species: K+, OH-, CH3NH3+, and Cl-

Yes a reaction goes to completion OH-(aq) + CH3NH3


OH- CH3NH3+ CH3NH2


Final (mol) 0 0.1 0.1

Major Species: K+, CH3NH2, Cl-, and CH3NH2

Buffer 47. a) Major Species: H+, NO3

-, Na+, and NO3-

No reaction goes to completion Not a buffer (strong acid) b) Major Species: H+, NO3

-, HF No reaction goes to completion

Not a buffer (strong acid and weak acid) c) Major Species: H+, NO3

-, Na+, and F- Yes a reaction goes to completion H+(aq) + F-(aq) HF(aq)

H+ F- HF


Final (mol) 0 0.2 0.2

Major Species: NO3-, Na+, and F-

Buffer d) Major Species: H+, NO3

-, Na+, and OH- Yes a reaction goes to completion H+(aq) + OH-(aq) H2O(aq)

H+ OH-

Initial (mol) 0.2 0.4

Final (mol) 0 0.2

Major Species: NO3-, Na+, and OH-

Not a buffer (strong base) 48. Major Species: Na+, F-, H+, and Cl-

A reaction goes to completion H+ + F- HF

H+ F- HF

Initial (mol) 0.0025 0.0100 0

Final (mol) 0 0.0075 0.0025

Major Species: Na+, F-, Cl-, and HF Buffer

𝐾𝑎 = 7.2 × 10−4 (Table 7.2)


[𝐻𝐴]) = −𝑙𝑜𝑔(7.2 × 10−4) + 𝑙𝑜𝑔 (

0.0075 𝑚𝑜𝑙𝑉

0.0025 𝑚𝑜𝑙𝑉

) = 3.62

12

49. A buffer has the greatest buffer capacity when [HA] = [A-]. When this happens the pH = pKa. Therefore, the best acid would have a pKa of 7.00. The acid that has the closest pKa value is HOCl which has a pKa of 7.46. In order to make a 1 L solution you would mix equal amounts of HOCl with NaOCl (or another salt containing OCl-).

54. a) The blue line is the weak acid and the red line is the strong acid. You can tell the

difference between the two plots because of the following 3 reasons. 1) The equivalence point of a weak acids pH curve is at a pH great than 7 and the equivalence point of a strong acid pH curve is at 7. 2) At the start of the pH curve of a weak acid titration the pH changes quickly and then levels off. This does not happen in a strong acid titration curve. 3) Since both acids have the same initial concentration the starting point of the strong acid should be at a lower pH than the weak acid.

b) While the definition of a buffer solution is a solution that resist change in pH a better definition would be a solution in which the H+ and OH- concentrations are constant. In the weak acid/strong base titration, as the strong base is added to the weak acid the weak acid is converted into its conjugate base. When there is equal amount of the weak acid and weak base this is called the half equivalence point. At this point if an acid is added the weak base can react with the acid to absorb the added H+ in the solution resulting in approximately the same amount of H+. If a base is added the weak acid can react with the base to absorb the added OH- in the solution resulting in approximately the same OH-. Therefore, the middle of the buffer region sits at the half equivalence point. Although the pH is stable at the half equivalence point for the strong acid/strong base titration this system is not considered a buffer because at very low pH’s a very small change in pH results in a large change in H+ concentration causing the concentration of the H+ and OH- to change. When strong base is added to the acid system the H+ reacts with the OH- to form water. However, water is neutral therefore, if an acid was added to the system there would be nothing to react with it causing the H+ concentration to go down dramatically.

c) The statement is true. Since both acid have the same initial amount and initial concentrations, the amount of base added to get to the equivalence point will be the same for both systems. The definition of the equivalence point is the point at which enough titrant has been added to fully react the analyte. Since the moles of acid is the same for both the strong and weak acid the number of moles of base needed to fully react each system is the same.

d) The statement is false. When a strong acid reacts with a strong base a neutral salt and water is formed. This results in a pH of 7.00 at the equivalence point. When a weak acid reacts with a strong base the conjugate base of the weak acid is formed. At the equivalence point all of the weak acid will be converted into the conjugate base of the acid resulting in a pH that is greater than 7.00 at the equivalence point.

55. a) The acid is a weak acid. You can tell this because at the beginning of the

titration (beaker c) all of the hydrogen atoms (blue atoms) are attached to their counter ions (green atoms). As strong base is added, hydrogen atoms are removed from their counter ions forming water and leaving behind the conjugate base (or counter ion) of the acid. If the system was a strong acid you

13

would see only hydrogen ions (blue atoms) in the solution at the beginning of the titration. As strong base is added, the hydrogen atoms should disappear because they are reacting with OH- to form water which is not shown in the picture.

b) c(start of titration) ae (half equivalence point)b (equivalence point)d(end of titration)

c) The pH=pKa for beaker e. This is the half equivalence point or the point where half of the weak acid (green atom bonded to blue atom) is converted to the conjugate base (green atom) of the weak acid.

d) Beaker b represents the equivalence point. The equivalence point occurs when all of the weak acid (green atom bonded to blue atom) is converted to the conjugate base (green atom) of the weak acid.

e) You would not need to know the Ka value of the weak acid to determine the pH of the system for beaker d. At this point all of the weak acid has been converted into the conjugate base of the weak acid. In addition, more strong base has been added. In a system with a strong base and a weak base the strong base determines the pH of the system due to the dramatically larger number of OH- ions from the strong base.

56. a) The equivalence point occurs when there are equal mole of weak acid and

strong base. For this plot it occurs after ~22 mL of base is added. The pH at the equivalence point will be greater than 7.

b) & c) The maximum buffering occurs when the [HA] = [A-] or pH = pKa. This will be in the middle of the flat region (~12).

d) The pH only depends on [HA] on the far left of the plot because no [A-] is present.

e) The pH only depends on [A-] at the equivalence point, when ~22 mL of base are added

f) The pH depends only on the amount of excess base on the far right of the plot.

14

57.

For weak base titrated with a strong acid the equivalence point should occur at a pH lower than 7 due to the weak conjugate acid that is generated.

61. a) f. The pH initially increases more rapidly for weaker acids than it does for

stronger acids. In, addition the weaker the acid the stronger the conjugate base therefore the weakest acid should have the highest pH at the equivalence point.

b) a. If you did not know the initial concentration of the acids and wanted to know if the acid was strong or weak you could look at the pH at the equivalence point.

If the pH at the equivalence point is 7.0 then it is a strong acid if the pH at the equivalence point is greater than 7.0 it is a weak acid.

c) d. The pKa of the system is 6.0, therefore, when [HA] = [A-] the pH of the system should be 6.0. This is referred to as the ½ equivalence point because ½ the amount of base has been added that would be needed to get to the equivalence point. For this graph the ½ equivalence point occurs after ~25 mL of NaOH has been added. The line that has a pH closest to 6 at the ½ equivalence point is d.

63. This is a strong acid strong base titration Calculate the initial moles of HClO4

0.0400 𝐿 𝐻𝐶𝑙𝑂4 (0.200 𝑚𝑜𝑙 𝐻𝐶𝑙𝑂4

1 𝐿 𝐻𝐶𝑙𝑂4) (

1 𝑚𝑜𝑙 𝐻+

1 𝑚𝑜𝑙 𝐻𝐶𝑙𝑂4) = 0.00800 𝑚𝑜𝑙 𝐻+

a) Major Species: H+, and ClO4-

No reaction goes to completion Not a buffer

𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔 (0.00800 𝑚𝑜𝑙

0.0400 𝐿) = 0.699

b) Major Species: H+, ClO4-, K+, and OH-

A reaction goes to completion H+(aq) + OH-(aq) H2O(l)

H+ OH-

Initial (mol) 0.00800 0.00100

Final (mol) 0.00700 0

Major Species: H+, ClO4-, and K+

Not a buffer

𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔 (0.00700 𝑚𝑜𝑙

0.0400 𝐿 + 0.0100 𝐿) = 0.854

15

c) Major Species: H+, ClO4-, K+, and OH-


H+ OH-

Initial (mol) 0.00800 0.00400

Final (mol) 0.00400 0

Major Species: H+, ClO4-, and K+

Not a buffer

𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔 (0.00400 𝑚𝑜𝑙

0.0400 𝐿 + 0.0400 𝐿) = 1.301

d) Major Species: H+, ClO4-, K+, and OH-


H+ OH-

Initial (mol) 0.00800 0.00800

Final (mol) 0 0

Major Species: ClO4-, and K+

Not a buffer Since the moles of OH- equal the moles of H+ the solution will be neutral 7.00. Another way to state this is that we are at the equivalence point.

e) Major Species: H+, ClO4-, K+, and OH-


H+ OH-

Initial (mol) 0.00800 0.01000

Final (mol) 0 0.00200

Major Species: ClO4-, K+, and OH-

Not a buffer

𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻−] = −𝑙𝑜𝑔 (0.00200 𝑚𝑜𝑙

0.0400 𝐿 + 0.100 𝐿) = 1.85

𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.00 − 1.85 = 12.15 64. Ba(OH)2 is a strong base and HCl is a strong acid Calculate the moles of Ba(OH)2 that you start with

0.0800 𝐿 𝐵𝑎(𝑂𝐻)2 (0.100 𝑚𝑜𝑙 𝐵𝑎(𝑂𝐻)2

1 𝐿 𝐵𝑎(𝑂𝐻)2) (

2 𝑚𝑜𝑙 𝑂𝐻−

1 𝑚𝑜𝑙 𝐵𝑎(𝑂𝐻)2) = 0.0160 𝑚𝑜𝑙 𝑂𝐻−

a) Major Species: Ba2+, and OH- No reaction goes to completion Not a buffer


0.0800 𝐿) = 0.699

𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.000 − 0.699 = 13.301

16

b) Major Species: Ba2+, OH-, H+, and Cl- Yes a reaction goes to completion H+(aq) + OH-(aq) H2O(l)

OH- H+

Initial (mol) 0.0160 0.00800

Final (mol) 0.0080 0

Major Species: Ba2+, OH-, and Cl- Not a buffer


0.0800 𝐿 + 0.0200 𝐿) = 1.10

𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.00 − 1.10 = 12.390 c) Major Species: Ba2+, OH-, H+, and Cl-

Yes a reaction goes to completion H+(aq) + OH-(aq) H2O(l)

OH- H+

Initial (mol) 0.0160 0.0120

Final (mol) 0.0040 0

Major Species: Ba2+, OH-, and Cl- Not a buffer


0.0800 𝐿 + 0.0300 𝐿) = 1.44

𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.00 − 1.44 = 12.56 d) Major Species: Ba2+, OH-, H+, and Cl-

Yes a reaction goes to completion H+(aq) + OH-(aq) H2O(l)

OH- H+

Initial (mol) 0.0160 0.0160

Final (mol) 0 0

Major Species: Ba2+, and Cl-

Not a buffer Since the moles of OH- equal the moles of H+ the solution will be neutral 7.00. Another way to state this is that we are at the equivalence point.

e) Major Species: Ba2+, OH-, H+, and Cl- Yes a reaction goes to completion H+(aq) + OH-(aq) H2O(l)

OH- H+

Initial (mol) 0.0160 0.0320

Final (mol) 0 0.0160

Major Species: Ba2+, H+, and Cl-

Not a buffer

𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔 (0.0160 𝑚𝑜𝑙

0.0800 𝐿 + 0.0800 𝐿) = 1.000

65. This is a weak acid/strong base titration

a) Major Species: HC2H3O2 No reaction goes to completion Not a buffer

17


HC2H3O2 H+ C2H3O2-

Initial (M) 0.200 0 0

Change (M) -x +x +x


𝐾𝑎 =[𝐻𝐶2𝐻3𝑂2]

[𝐻+][𝐶2𝐻3𝑂2−]

=𝑥𝑥

(0.200 − 𝑥)= 1.8 × 10−5

Since Ka is small assume 0.200-x = 0.200 𝑥𝑥

(0.200)= 1.8 × 10−5

𝑥 = 0.0019 Check assumption

0.0019

0.200100% = 0.95% Good

Calculate pH [𝐻+] = 𝑥 = 0.0019 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(0.0019) = 2.72

b) Major Species: HC2H3O2, K+, and OH- A reaction goes to completion

HC2H3O2(aq) + OH-(aq) H2O(l) + C2H3O2-(aq)

Calculate Initial moles of HC2H3O2 and KOH

0.100 𝐿 𝐻𝐶2𝐻3𝑂2 (0.200 𝑚𝑜𝑙 𝐻𝐶2𝐻3𝑂21 𝐿 𝐻𝐶2𝐻3𝑂2

) = 0.0200 𝑚𝑜𝑙 𝐻𝐶2𝐻3𝑂2

0.1000 𝐿 𝐾𝑂𝐻 (0.100 𝑚𝑜𝑙 𝐾𝑂𝐻

1 𝐿 𝐾𝑂𝐻) (

1 𝑚𝑜𝑙 𝑂𝐻−

1 𝑚𝑜𝑙 𝐾𝑂𝐻) = 0.00500 𝑚𝑜𝑙 𝑂𝐻−

HC2H3O2 OH- C2H3O2-

Initial (mol) 0.0200 0.00500 0

Final (mol) 0.0150 0 0.00500

Major Species: HC2H3O2, K+, and C2H3O2-

Buffer


[𝐻𝐴]) = −𝑙𝑜𝑔(1.8 × 10−5) + 𝑙𝑜𝑔 (

0.00500 𝑚𝑜𝑙0.1500 𝐿

0.0150 𝑚𝑜𝑙0.1500 𝐿

)

= 4.26

c) Major Species: HC2H3O2, K+, and OH- A reaction goes to completion


HC2H3O2 OH- C2H3O2-

Initial (mol) 0.0200 0.0100 0

Final (mol) 0.0100 0 0.0100


Buffer


[𝐻𝐴]) = −𝑙𝑜𝑔(1.8 × 10−5) + 𝑙𝑜𝑔 (

0.0100 𝑚𝑜𝑙0.200 𝐿

0.0100 𝑚𝑜𝑙0.200 𝐿

)

= 4.74

18

d) Major Species: HC2H3O2, K+, and OH- A reaction goes to completion


HC2H3O2 OH- C2H3O2-

Initial (mol) 0.0200 0.0150 0

Final (mol) 0.0050 0 0.0150


Buffer


[𝐻𝐴]) = −𝑙𝑜𝑔(1.8 × 10−5) + 𝑙𝑜𝑔 (

0.01500 𝑚𝑜𝑙0.2500 𝐿

0.0050 𝑚𝑜𝑙0.2500 𝐿

)

= 5.22

e) Major Species: HC2H3O2, K+, and OH- A reaction goes to completion


HC2H3O2 OH- C2H3O2-

Initial (mol) 0.0200 0.0200 0

Final (mol) 0 0 0.0200

Major Species: C2H3O2-, and K+

Not a buffer C2H3O2

-(aq) + H2O(l) ⇌ HC2H3O2(aq) + OH-(aq) Calculate Kb

𝐾𝑏 =𝐾𝑤

𝐾𝑎=

1.0×10−14

1.8×10−5 = 5.6 × 10−10

C2H3O2- HC2H3O2 OH-

Initial (mol) 0.02000 0 0

Initial (M) 0.0667 0 0

Change (M) -x +x +x


𝐾𝑏 =[𝐶2𝐻3𝑂2

−]

[𝐻𝐶2𝐻3𝑂2][𝑂𝐻−]= 5.6 × 10−10 =

𝑥𝑥

(0.0667 − 𝑥)

Since Kb is very small assume 0.0667-x = 0.06667 𝑥𝑥

(0.0667 − 𝑥)= 5.6 × 10−10

𝑥 = 6.1 × 10−6 Check Assumption

6.1×10−6

0.0667100% = 0.0091% Good

Calculate pH [𝑂𝐻−] = 𝑥 = 6.1 × 10−6 𝑀 𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻−] = −𝑙𝑜𝑔(6.1 × 10−6) = 5.21 𝑝𝐻 + 𝑝𝑂𝐻 = 14.00 𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.00 − 5.21 = 8.79

19

f) Major Species: HC2H3O2, K+, and OH- A reaction goes to completion


HC2H3O2 OH- C2H3O2-

Initial (mol) 0.0200 0.0250 0

Final (mol) 0 0.0050 0.0200

Major Species: C2H3O2-, K+, and OH-

The majority of the OH- will come from OH-

Determine [OH-]

𝑀 =𝑛

𝑉=

0.0050 𝑚𝑜𝑙

0.3500 𝐿= 0.0143 𝑀

Calculate pH 𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻−] = −𝑙𝑜𝑔(0.00143) = 1.84 𝑝𝐻 + 𝑝𝑂𝐻 = 14.00 𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.00 − 1.84 = 12.16

66. This is a weak base/strong acid titration

a) Major Species: H2NNH2

No reaction goes to completion. Not a buffer

H2NNH2(aq) + H2O(l) ⇌ H2NNH3+(aq) + OH-(aq) 𝐾𝑏 = 3.0 × 10−6

H2NNH2 H2NNH3+ OH-

Initial (M) 0.100 0 0

Change (M) -x +x +x


𝐾𝑏 =[𝐻2𝑁𝑁𝐻3

+][𝑂𝐻−]

[𝐻2𝑁𝑁𝐻2]=

𝑥𝑥

(0.100 − 𝑥)= 3.0 × 10−6


(0.100 − 𝑥)= 3.0 × 10−6


5.5×10−4

0.100100% = 0.55% Good

Calculate pH [𝑂𝐻−] = 𝑥 = 5.5 × 10−4 𝑀 𝑝𝑂𝐻 = −𝑙𝑜𝑔[𝑂𝐻−] = −𝑙𝑜𝑔(5.5 × 10−4) = 3.26 𝑝𝐻 + 𝑝𝑂𝐻 = 14.00 𝑝𝐻 = 14.00 − 𝑝𝑂𝐻 = 14.00 − 3.26 = 10.74

b) Major Species: H2NNH2, H+, and NO3-

A reaction goes to completion H2NNH2(aq) + H+(aq) H2NNH3

+(aq)

H2NNH2 H+ H2NNH3+

Initial (mol) 0.0100 0.00400 0

Final (mol) 0.0060 0 0.00400

Major Species: H2NNH2, NO3-, and H2NNH3

+ Buffer

20

Determine the Ka

𝐾𝑎 =𝐾𝑤

𝐾𝑏=

1.0×10−14

3.0×10−6= 3.3 × 10−9


[𝐻𝐴]) = −𝑙𝑜𝑔(3.3 × 10−9) + 𝑙𝑜𝑔 (

0.0060 𝑚𝑜𝑙0.1200 𝐿

0.00400 𝑚𝑜𝑙0.1200 𝐿

)

= 8.66 c) Major Species: H2NNH2, H+, and NO3

- A reaction goes to completion

H2NNH2(aq) + H+(aq) H2NNH3+(aq)

H2NNH2 H+ H2NNH3+

Initial (mol) 0.0100 0.00500 0

Final (mol) 0.0050 0 0.00500


+ Buffer


[𝐻𝐴]) = −𝑙𝑜𝑔(3.3 × 10−9) + 𝑙𝑜𝑔 (

0.0050 𝑚𝑜𝑙0.1250 𝐿

0.00500 𝑚𝑜𝑙0.1250 𝐿

)

= 8.48 d) Major Species: H2NNH2, H+, and NO3



H2NNH2 H+ H2NNH3+

Initial (mol) 0.0100 0.00800 0

Final (mol) 0.0020 0 0.00800


+ Buffer


[𝐻𝐴]) = −𝑙𝑜𝑔(3.3 × 10−9) + 𝑙𝑜𝑔 (

0.0020 𝑚𝑜𝑙0.1400 𝐿

0.00800 𝑚𝑜𝑙0.1400 𝐿

)

= 7.88 e) Major Species: H2NNH2, H+, and NO3



H2NNH2 H+ H2NNH3+

Initial (mol) 0.0100 mol 0.0100 mol 0

Final (mol) 0 0 0.0100 mol

Major Species: H2NNH3+, and NO3

-

Not a buffer

H2NNH3+(aq) ⇌ H2NNH2(aq) + H+(aq) 𝐾𝑎 = 3.3 × 10−9

H2NNH3+ H2NNH2 H+

Initial (mol) 0.0100 0 0

Initial (M) 0.0667 0 0

Change (M) -x +x +x


21

𝐾𝑎 =[𝐻2𝑁𝑁𝐻2][𝐻+]

[𝐻2𝑁𝑁𝐻3+]

=𝑥𝑥

(0.0667 − 𝑥)= 3.3 × 10−9


(0.0667 − 𝑥)= 3.3 × 10−9


1.5×10−5

0.0667100% = 0.022% Good

Calculate pH [𝐻+] = 𝑥 = 1.5 × 10−5 𝑀 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔(1.5 × 10−5 ) = 4.82

f) Major Species: H2NNH2, H+, and NO3-

A reaction goes to completion H2NNH2(aq) + H+(aq) H2NNH3

+(aq)

H2NNH2 H+ H2NNH3+

Initial (mol) 0.0100 0.0200 0

Final (mol) 0 0.0100 0.0100

Major Species: H2NNH3+, H+, and NO3

- Not a buffer

The H+ concentration will be mainly from the excess strong acid.

𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] = −𝑙𝑜𝑔 (0.0100 𝑚𝑜𝑙

0.2000 𝐿) = 1.301

99. a) CaC2O4(s) ⇌ Ca2+(aq) + C2O4

2-(aq) 𝐾𝑠𝑝 = [𝐶𝑎2+][𝐶2𝑂4

2−]

Ca2+ C2O42-

Initial (M) 0 0

Change (M) +x +x

Equilibrium (M) x x

* x is the solubility The concentrations of Ca2+ and C2O4

2- must be in M.

(6.1×10−3 𝑔 𝐶𝑎𝐶2𝑂4

1 𝐿 𝐶𝑎𝐶2𝑂4) (

1 𝑚𝑜𝑙 𝐶𝑎𝐶2𝑂4

128.10 𝑔 𝐶𝑎𝐶2𝑂4) = 4.8 × 10−5 𝑀

𝐾𝑠𝑝 = [𝐶𝑎2+][𝐶2𝑂42−] = 𝑥𝑥 = 𝑥2 = (4.8 × 10−5)2 = 2.3 × 10−9

b) BiI3(s) ⇌ Bi3+(aq) + 3I-(aq) 𝐾𝑠𝑝 = [𝐵𝑖3+][𝐼−]3

Bi3+ I-

Initial (M) 0 0

Change (M) +x +3x

Equilibrium (M) x 3x

* x is the solubility 𝐾𝑠𝑝 = [𝐵𝑖3+][𝐼−]3 = 𝑥(3𝑥)3 = 27𝑥4 = 27(1.32 × 10−5)4 = 8.20 × 10−19

22

106. a) Ag2SO4(s) ⇌ 2Ag+(aq) + SO42-(aq)

𝐾𝑠𝑝 = [𝐴𝑔+]2[𝑆𝑂42−]

Ag+ SO42-

Initial (M) 0 0

Change (M) +2x +x

Equilibrium (M) 2x x

𝐾𝑠𝑝 = [𝐴𝑔+]2[𝑆𝑂42−] = (2𝑥)2𝑥 = 4𝑥3 = 1.2 × 10−5

𝑥 = 0.014 𝑀 b) Ag2SO4(s) ⇌ 2Ag+(aq) + SO4

2-(aq) 𝐾𝑠𝑝 = [𝐴𝑔+]2[𝑆𝑂4

2−]

Ag+ SO42-

Initial (M) 0.10 0

Change (M) +2x +x

Equilibrium (M) 0.10+2x x

Assume that x is small due to the small equilibrium constant 𝐾𝑠𝑝 = [𝐴𝑔+]2[𝑆𝑂4

2−] = (0.10 + 2𝑥)2𝑥 = 1.2 × 10−5

Assume that 0.10 +2x =0.10 (0.10)2𝑥 = 1.2 × 10−5 𝑥 = 0.0012 𝑀

Check assumption 2(0.0012)

0.10100% = 2.4% Good

c) Ag2SO4(s) ⇌ 2Ag+(aq) + SO42-(aq)

𝐾𝑠𝑝 = [𝐴𝑔+]2[𝑆𝑂42−]

Ag+ SO42-

Initial (M) 0 0.20 M

Change (M) +2x +x

Equilibrium (M) 2x 0.20+x

Assume that x is small due to the small equilibrium constant 𝐾𝑠𝑝 = [𝐴𝑔+]2[𝑆𝑂4

2−] = (2𝑥)2(0.20 + 𝑥) = 1.2 × 10−5

Assume that 0.20 +x =0.20 (2𝑥)2(0.20) = 1.2 × 10−5 𝑥 = 0.0039 𝑀

Check assumption 0.0039

0.20100% = 2.0% Good

107. a) Fe(OH)3(s) ⇌ Fe3+(aq) + 3OH-(aq)

𝐾𝑠𝑝 = [𝐹𝑒3+][𝑂𝐻−]3

This problem is slightly different because there is some OH- in water. From the Kw we know that the initial [OH-]=1.0×10-7.

Fe3+ OH-

Initial (M) 0 1×10-7

Change (M) +x +3x

Equilibrium (M) x 1×10-7+3x

* x is the solubility

23

Since Ksp is small assume that 1×10-7+3x = 1×10-7 𝐾𝑠𝑝 = [𝐹𝑒3+][𝑂𝐻−]3 = 𝑥(1 × 10−7)3 = 4 × 10−38

𝑥 = 4 × 10−17 𝑀 Check assumption

3(4×10−17)

1×10−7 100% = 1 × 10−7% Good

b) Fe(OH)3(s) ⇌ Fe3+(aq) + 3OH-(aq) 𝐾𝑠𝑝 = [𝐹𝑒3+][𝑂𝐻−]3

Calculate the concentration of OH- ions when the pH is 5.0 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] [𝐻+] = 10−𝑝𝐻 = 10−5.0 = 1 × 10−5 [𝐻+][𝑂𝐻−] = 1 × 10−14

[𝑂𝐻−] =1 × 10−14

[𝐻+]=

1 × 10−14

1 × 10−5= 1 × 10−9

Fe3+ OH-

Initial (M) 0 1×10-9

Change (M) +x +3x

Equilibrium (M) x 1×10-9*

* x is the solubility * The question told you to assume that the pH is constant

𝐾𝑠𝑝 = [𝐹𝑒3+][𝑂𝐻−]3 = 𝑥(1 × 10−9)3 = 4 × 10−38

𝑥 = 4 × 10−11 𝑀 c) Fe(OH)3(s) ⇌ Fe3+(aq) + 3OH-(aq) 𝐾𝑠𝑝 = [𝐹𝑒3+][𝑂𝐻−]3

Calculate the concentration of OH- ions when the pH is 11.0 𝑝𝐻 = −𝑙𝑜𝑔[𝐻+] [𝐻+] = 10−𝑝𝐻 = 10−11.0 = 1 × 10−11 [𝐻+][𝑂𝐻−] = 1 × 10−14

[𝑂𝐻−] =1 × 10−14

[𝐻+]=

1 × 10−14

1 × 10−11= 0.001

Fe3+ OH-

Initial (M) 0 0.001

Change (M) +x +3x

Equilibrium (M) x 0.001*

* x is the solubility * The question told you to assume that the pH is constant

Since Ksp is small assume that 0.001 + 3x = 0.001 𝐾𝑠𝑝 = [𝐹𝑒3+][𝑂𝐻−]3 = 𝑥(0.001)3 = 4 × 10−38

𝑥 = 4 × 10−29 𝑀

homework #3 chapter 8 · 1 homework #3 chapter 8 applications of aqueous equilibrium 15. buffer...

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