BIT Chapters 17-20

1. Kw = [H+][OH–] = 9.5 × 10–14

[H+] = [OH–] = 3.1 × 10–7

pH = 6.51

The solution is neither acidic nor basic because the concentration of the hydronium ion equals the

concentration of the hydroxide ion.

2. The concentration of OH– is 4.7 10–7 g L–1

The molar concentration of OH– is:

[OH–] = 7 — —

—

4.7 10 g OH 1 mol OH

1 L solution 17.01 g OH = 2.76 10–8 M

pOH = –log[OH–] = –log(2.76 10–8 M) = 7.56 pH = 14 – pOH = 14 – 7.56 = 6.44

The water is acidic.

3. (a) C6H5OH(aq) + H2O(l) C6H5O–(aq) + H3O

+(aq)

(b) Ka = 6 5 3

6 5

C H O H O

C H OH

(c) [C6H5OH] = 0.550 M since the amount of dissociation is negligible

[H3O+] = –log(pH) = –log(5.07) = 8.51 10–6 M

[C6H5O–] = 8.51 10–6 M since all of the H3O

+ comes from the dissociation of the phenol

Ka = 6 5 3

6 5

C H O H O

C H OH =

6 68.51 10 8.51 10

0.550 = 1.3 10–10

pKa = –log Ka = –log(1.3 10–10) = 9.89

(d) Kb = 14 14

10a

1 10 1 10

K 1.3 10 = 7.69 10–5

pKb = –log Kb = –log(7.69 10–5) = 4.11

4. (a) pKb = 14 – pKa = 14 – 11.68 = 2.32

(b) C7H3SO3– + H2O C7H3SO3H + OH–

The solution will be basic

Kb = 7 3 3

7 3 3

C H SO H OH

C H SO = 4.79 10–3

[C7H3SO3–] [C7H3SO3H] [OH

–]

I 0.010 – –

C –x +x +x

E 0.010–x +x +x

4.79 10–3 = x x

0.010 x

x will be large compared to 0.010 M. Therefore we need to solve it either by successive

approximations or the quadratic equation.

x = 4.92 10–3

= [OH–]

pOH = 2.31

pH = 11.69

BIT Chapters 17-20

5. Let Cod stand for codeine

Cod + H2O CodH+ + OH–

Kb = CodH OH

Cod = 1.63 10

–6

[Cod] [CodH+] [OH

–]

I 0.0115 – –

C –x +x +x

E 0.0115–x x x

1.63 10–6 = x x

0.0115 x

x is large compared ot 0.0115, therefore solve for x using either the quadratic equation or successive

approximations

x = 1.37 10–4 = [OH–] pOH = 3.86

pH = 14 – pH = 10.14

6. CH3NH2(aq) + H2O CH3NH3+(aq) + OH–(aq)

Kb = 3 3

3 2

CH NH OH

CH NH

7. pKb = 3.36

pKa of its conjugate acid = 14 – pKb = 14 – 3.36 = 10.64

8. (a) M ascorbic acid solution = 2 6 6 6 2 6 6 6

2 6 6 6

3.12 g H C H O 1 mol H C H O

0.125 L 176.1 g H C H O = 0.142 M H2C6H6O6

(b) H2C6H6O6 HC6H6O6– + H3O

+ Ka1 = 6 6 6 3

2 6 6 6

HC H O H O

H C H O = 8.0 10–5

HC6H6O6– C6H6O6

2– + H3O+ Ka2 =

26 6 6 3

6 6 6

C H O H O

HC H O = 1.6 10–12

Since all of the H3O+ will come from the first equilibrium reaction, it can be calculated as follows:

[H2C6H6O6] [HC6H6O6–] [H3O

+]

I 0.142 – –

C –x +x +x

E 0.142–x x x

8.0 10–5 =

6 6 6 3

2 6 6 6

HC H O H O

H C H O

8.0 10–5 =

x x

0.142 x

Solve for x

x = 3.37 10–3 = [H3O+]

pH = –log[H3O+] = –log(3.37 10–3) = 2.47

[C6H6O62–] = 1.6 10–12 [HC6H6O6

–] = [H3O+] in the Ka2 equation, these cancel

BIT Chapters 17-20

9. pH = pKa + log initial

initial

A

HA

4.50 = 4.74 + log initial

initial

A

HA

–0.24 = log initial

initial

A

HA

10–0.24 = initial

initial

A

HA

initial

initial

A

HA = 0.575

10. (a) C2H3O2

– + H+ HC2H3O2

(b) HC2H3O2 + OH– C2H3O2– + H2O

11. Ka = 1.8 10–5 = 2 3 2 3

2 3 2

C H O H O

HC H O

30.10 H O

0.15 = 1.8 10–5

[H3O+] = 2.7 10–5

pH = 4.57

[HC2H3O2]final = (0.15 – 0.04)M = 0.11 M

[C2H3O2–]final = (0.10 + 0.04)M = 0.14 M

30.14 H O

0.11= 1.8 10–5

[H3O+] = 1.4 10–5

pH = 4.85

The change in pH is 0.28 pH units.

12. (a) Propanoic acid, pKa = 4.89 and its salt, sodium propionate

(b) This problem is best solved using two equations and two unknowns:

pH = pKa + logmol A

mol HA

5.10 = 4.89 + logA 0.005

HA 0.005

5.00 = 4.89 + logA

HA

BIT Chapters 17-20

0.21 = logA 0.005

HA 0.005

0.11 = logA

HA

100.21 = A 0.005

HA 0.005

100.11 = A

HA

[A–] = 100.11[HA]

100.21 =

0.1110 HA 0.005

HA 0.005

100.21([HA] – 0.005) = 100.11[HA] + 0.005

1.622[HA] – 0.008109 = 1.288[HA] + 0.005

0.334[HA] = 0.01311 [HA] = 0.03925 M

[A–] = 100.11(0.03925 M) = 0.05056 M

For 625 mL (0.625 L) of solution:

For determining the mass of propionic acid needed:

g C3H6O2 = 0.625 L 3 6 2 3 6 2

3 6 2

0.03925 mol C H O 74.09 g C H O

1 L 1 mol C H O = 1.82 g C3H6O2

For determining the mass of sodium propionate:

g NaC3H5O2 = 0.625 L 3 5 2 3 5 2

3 5 2

0.05056 mol NaC H O 96.07 g NaC H O

1 L 1 mol NaC H O = 3.04 g

NaC3H5O2

(c) From the above calculation:

The concentration of propionic acid is 0.03925 M.

The concentration of sodium propionate is 0.05056 M.

13. (a) Neutral

(b) Acidic

(c) Acidic

(d) Basic

14. pKb = 14 – pKa = 14 – 4.87 = 9.13

Kb = HA OH

A = 7.41 10–10

All of the acid has been converted into the salt at the equivalence point.

(0.115 M)(50.00 mL) = (0.100 M)(x mL)

x = 57.50 mL total volume is 107.50 mL

[A–] = 0.05349 M

[HA] = [OH–] = x

BIT Chapters 17-20

7.41 10–10 =x x

0.05349

x = 6.296 × 10–6 M = [OH–] pOH = 5.201

pH = 8.799

Thymol blue or phenolphthalein would be good indicators.

15. Kb1 =

2

14w

12a

K 1.0 10

K 1.6 10 = 6.3 10–3

[C6H6O62–

]

[HC6H6O6–] [OH

–]

I 0.050 – –

C –x +x +x

E 0.050–x x x

6 6 6

26 6 6

HC H O OH

C H O =

x x

0.050 x = 6.3 10

–3

x is too large to make a simplifying assumption, therefore solve for x either using the quadratic equation or

by successive approximations.

x = 0.015 = [OH–]

pOH = 1.82 pH = 12.18

16. Since half of the acid is neutralized the concentration of the acid is equal to the concentration of its

conjugate base, the pKa can be determined:

pH = pKa + logA

HA

3.56 = pKa + log 1

pKa = pH = 3.56

Ka = 10–pKa = 10–3.56 = 2.75 10–4

17. pH = pKa + logA

HA

pKa = 14 – pKb pKb = –log Kb = –log 1.8 10–5 pKa = 14 – 4.74 = 9.26

9.26 = 9.26 + logA

HA

[A–] = [HA]

[NH4+] = 0.100 before the addition of NaOH. In order for the two concentrations to be equal, half as much

NaOH must be added:

(0.100 L NH4+)(0.100 M NH4

+ solution) = 0.0100 mol NH4+

(0.0100 mol NH4+)(0.5) = 0.00500 mol NaOH

0.00500 mol NaOH40.00 g NaOH

1 mol NaOH = 0.200 g NaOH

18. Ksp = [Ag+]2[CrO42–]

[Ag+] = 2 (6.5 10–5 M) = 1.30 10–4 M

[CrO42–] = 6.5 10–5 M

Ksp = (1.30 10–4)2(6.5 10–5)

BIT Chapters 17-20

Ksp = 1.1 10–12

19. Mg(OH)2(s) Mg2+(aq) + OH–(aq)

Ksp = 5.6 10–12 = [Mg2+][OH–]2

5.6 10–12 = [x][2x]2

x = 1.1 10–4

2x = 2.2 10–4 = [OH–]

pOH = –log[OH–] = –log(2.2 10–4) = 3.66 pH = 14 – 3.66 = 10.34

20. Fe(OH)2(s) Fe2+(aq) + 2OH–(aq)

ksp = 4.9 10–17 = [Fe2+][OH–]2 pH = 10.00

pOH = 14 – 10.00 = 4.00

[OH–] = 10–4.00 = 1 10–4

4.9 10–17 = [Fe2+][1 10–4]2

4.9 10–9 M = [Fe2+]

g L–1 =

9 2+2 2

2+2

1 mol Fe(OH) 89.86 g Fe(OH)4.9 10 mol Fe

1 L solution 1 mol Fe(OH)1 mol Fe = 4.40 10–7 g L–1

21. (e)

22. (a) This is a limiting reagent problem.

mL KI needed = 3 23 2

1 mol Pb(NO )1.04 g Pb(NO )

331.2 g

3 2

2 mol KI 1000 mL KI solution

1 mol Pb(NO ) 0.500 mol KI = 12.6 mL KI solution

20.0 mL KI solution is supplied. KI is in excess.

g PbI2 = 3 23 2

1 mol Pb(NO )1.04 g Pb(NO )

331.2 g

2 2

3 2 2

1 mol PbI 461.0 g PbI

1 mol Pb(NO ) 1 mol PbI = 1.45 g PbI2

(b) The concentration of the spectator ions:

[K+] = (0.500 M K+)(20.0 mL solution) 50.0 mL = 0.200 M K+

[NO3–] = (0.100)(2)(30.0 mL solution) 50.0 mL = 0.12 M NO3

– [I–]: Total moles of I– = (0.500 M I–)(0.0200 L solution) = 0.0100 mol I–

mol I– used = 3 2

3 2 2+

1 mol Pb(NO ) 2 mol I1.04 g Pb(NO )

331.2 g 1 mol Pb = 6.3 10–3 mol I–

[I–] = (0.0100 mol I– – 6.3 10–3 mol I–)/(0.050 L solution) = 0.0740 M I– [Pb2+]: This will be the amount of Pb2+ that is in solution after the solid PbI2 reaches equilibrium

PbI2(s) Pb2+(aq) + 2I–(aq)

BIT Chapters 17-20

[Pb2+

]

[I–]

I – 0.0740 M

C +x +x

E x 0.0740 + x

Ksp = 9.8 10–9 = [Pb2+][I–]2 = (x)(0.0740 + x)2 x << 0.0740

9.8 10–9 = (x)(0.0740)2

x = 1.79 10–6 M = [Pb2+]

23. This is a simultaneous equilibrium problem.

CuCO3(s) Cu2+(aq) + CO32–(aq) Ksp = [Cu2+][CO3

2–] =2.5 10–10

Cu2+(aq) + 4NH3(aq) Cu(NH3)42+(aq) Kform =

2

3 4

423

Cu NH

Cu NH = 1.1 1013

CuCO3(s) + 4NH3(aq) Cu(NH3)42+(aq) + CO3

2–(aq)

Koverall =

2 23 34

4

3

Cu NH CO

NH = 2.75 103

All of the Cu(NH3)42+ comes from the CuCO3

[Cu(NH3)42+] = 1.00 g CuCO3

3

3

1 mol CuCO 1

123.6 g CuCO 1.00 L solution = 8.09 10–3

As the Cu2+ is used to form Cu(NH3)42+, the [CO3

2–] = [Cu(NH3)42+] = 8.09 10–3

2.75 103 =

2 23 34

4

3

Cu NH CO

NH =

3 3

4

8.09 10 8.09 10

x

x = 0.0124 M NH3

mol NH3 = 1.00 L solution 30.0124 mol NH

1 L solution =0.0124 mol NH3

24. Assume the solution is saturated with CO2 and therefore the concentration of H2CO3 is 0.030 M

H2CO3 2H+ + CO32–

K = Ka1 Ka

2 = (4.3 10–7)(5.6 10–11) = 2.4 10–17 =

22–

3

2 3

H CO

H CO

First, calculate the [H+] at which thePbCO3 will begin to precipitate:

PbCO3(s) Pb2+(aq) + CO32–(aq) Ksp = 7.4 10–14 = [Pb2+][CO3

2–]

[Pb2+] = 0.010 M

[CO32–] = (7.4 10–14)/(0.010) = 7.4 10–12 M

[H+] =

1 1

–17 –172 22 3

2– –123

2.4 10 H CO 2.4 10 0.030

CO 7.4 10 = 3.1 10–4

pH = 3.50

Now, determine the [H+] at which the BaCO3 will begin to precipitate:

BaCO3(s) Ba2+(aq) + CO32–(aq) Ksp = 2.6 10–9 = [Ba2+][CO3

2–]

BIT Chapters 17-20

[Ba2+] = 0.010 M

[CO32–] = (2.6 10–9)/(0.010) = 2.6 10–7 M

[H+] =

1 1

–17 –172 22 3

2– –73

2.4 10 H CO 2.4 10 0.030

CO 2.6 10 = 1.7 10–6

pH = 5.78

The pH range is between 3.50 and 5.78, above 5.78, BaCO3 will begin to precipitate.

25. The less soluble substance is PbS. We need to determine the minimum [H+] at which NiS will precipitate.

2+2

spa 2 + 2+

+

Ni H S (0.100)(0.1)K = = = 40 (from Table 18.2)

[H ]H

(0.10)(0.1)[H ] = = 0.016

40

pH = –log[H+] = 1.80. At a pH lower than 1.80, PbS will precipitate and NiS will not. At larger values of

pH, both PbS and NiS will precipitate.

We also need to determine the [H+] at which PbS will start to precipitate 2+

2 7spa 2 + 2

+

+

7

Pb H S (0.100)(0.1)K = = = 3 10 (from Table 18.2)

[H ]H

(0.10)(0.1)[H ] = = 182

3 10

pH = –log[H+] = –2.26. Any acid in water will precipitate the PbS.

The pH range is –2.26 to 1.80 to allow the PbS to precipitate without the NiS.

26. The less soluble substance is SnS. We need to determine the minimum [H2S] at which FeS will precipitate. 2+

2 2spa 2 3 2

+

3 23

2

Fe H S (0.10) H SK = = = 600 (from Table 18.2)

[1 10 ]H

(600)(1 10 )H S = = 6 10

(0.1)

The concentration of Sn2+ can now be determined.

2+ 2+ 32 5

spa 2 —3 2+

5 3 22+ 9

3

Sn H S Sn 6 10K = = = 1 10 (from Table 18.2)

[1 10 ]H

(1 10 )(1 10 )Sn = = 1.7 10

(6 10 )

27. (a) MS(s) M2+(aq) + S2–(aq)

Ksp = [M2+][S2–] = 4.0 × 10–29

MS(s) + H2O M2+(aq) + HS2–(aq) + OH–(aq)

BIT Chapters 17-20

Kspa =

22

2

M H S

H

= (4.0 × 10–29)(1021) = 4.0 × 10–8

(b) 2

2

x

0.30 = 4.0 × 10–8

x = 6.0 × 10–5 M

(c) MS would be considered an acid insoluble sulfide making M a group 2 ion. Group 3 ions form

insoluble sulfides in base.

28. 3NO(g) NO2(g) + N2O(g)

G° = {1 mol G°f[NO2(g)] + 1 mol G°f[N2O(g)]} – {3 mol G°f[NO(g)]}

G° = {1 mol (51.84 kJ/mol) + 1 mol (103.6 kJ/mol)} – {3 mol (–86.69 kJ/mol)}

G° = 416 kJ

G° = RTlnKP

4.16 105 J = –(8.314 J mol–1 K–1)(298 K)(ln KP) ln KP = –168

KP = 1.2 10–73

KP = Kc(RT) ng

1.2 10–73 = Kc[(0.0821 L atm mol–1 K–1)(298)]–1

Kc = 2.9 10–72

29. CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)

oΔG = {

ofΔG [CH3Cl(g)] +

ofΔG [HCl(g)]} – {

ofΔG [CH4(g)] +

ofΔG [Cl2(g)]}

oΔG = {1 mol (–58.6 kJ/mol) + 1 mol (–95.27 kJ/mol)}

– {1 mol (–50.79 kJ/mol) + 1 mol (0 kJ/mol)} oΔG = –103.08 kJ = 1.03 105 J

G° = RTlnKP

–1.03 105 J = – (8.314 J mol–1 K–1)(473 K)(ln KP) ln KP = 26.19

KP = 2.4 1011

KP = Kc(RT) ng

2.4 1011 = Kc[(0.0821 L atm mol–1 K–1)(473)]0

Kc = 2.4 1011

3 11

4 2

CH Cl HClK 2.4 x 10

CH Clc

4

1 mol4.56 g

16.04 gCH 0.142

2.00 LM

BIT Chapters 17-20

2

1 mol8.67 g

70.91 gCl 0.0611

2.00 LM

Some CH4 is lost to form CH3Cl and some Cl2 is lost to form HCl.

[CH4] [Cl2] [CH3Cl] [HCl]

I 0.142 0.0611 – C –x x +x +x

E 0.142 – x 0.0611 x +x +x

11

c

x xK 2.4 x 10

0.142 x 0.0611 x

The equilibrium lies so far to the right that you do not need to solve the equation. Cl2 is the limiting reagent

and thus will be completely consumued in the reaction.

At equilibrium the following concentrations exist:

[Cl2] = 0 ( actually it is slightly greater than zero but extremely small)

[CH4] = (2.0 L x 0.142 M 2.0 L x 0.0611 M)/2.0 L = 0.0809 M

[CH3Cl] = [ HCl] = 0.0611 M

30. First calculate the atomization energy for the formation of C2H6, and the atomization energy for the C2H2

and H2. The difference is the energy required for the reaction. Then calculate the amount of heat required

for 25.0 g of C2H6.

atomization energy1 = (6 mol C–H B.E.) + (1 mol C–C B.E.)

= (6 mol 412 kJ mol–1) + (1 mol 348 kJ mol–1) = 2820 kJ

atomization energy2 = (2 mol C–H B.E.) + (1 mol C C B.E.) + (1 mol H–H B.E.)

= (2 mol 412 kJ mol–1) + (1 mol 960 kJ mol–1) + (1 mol 436 kJ mol–1) = 2220 kJ

atomization energy1 – atomization energy2 = 2820 kJ – 2220 kJ = 600 kJ

600 kJ are absorbed.

kJ for 25.0 g = 25.0 g 2 6

2 6 2 6

1 mol C H 600 kJ

26.04 g C H 1 mol C H = 576 kJ

31. (a) mole e– = 20.0 min4

60 s 1.00 C 1 mol e

1 min 1 s 9.65 10 C = 1.24 10–2 mol e–

mole OH– = 1.24 10–2 mol e–1 mol OH

1 mol e = 1.24 10–2 mol OH–

[OH–] = 21.24 10 mol OH

0.250 L NaCl solution = 0.0497 M OH–

pOH = –log[OH–] = 1.303

pH = 14 – pOH

pH = 14 – 1.303 = 12.697

BIT Chapters 17-20

(b) mole e– = 10.0 min4

60 s 5.00 C 1 mol e

1 min 1 s 9.65 10 C = 3.11 10–2 mol e–

mole H2 = 3.11 10–2 mol e–21 mol H

2 mol e = 1.56 10–2 mol H2

mL H2 =

2 1 121.56 10 mol H 0.0821 L atm mol K 273 K 1000 mL

1 atm 1 L = 348 mL

32. (a)

(b) Fe(s) + Cu2+(aq) Cu(s) + Fe2+(aq)

(c) Fe|Fe2+||Cu2+|Cu

(d) E cell = E substance reduced – E substance oxidized

2+ 2

o o ocell Cu Fe

E E E

ocellE = 0.34 V – (–0.44 V) = 0.78 V

(e) e– = 50.0 h

–3600 s 0.10 C 1 mol e

1 h s 96,500 C = 1.87 10

–1 mol e

–

mol Fe2+ = 1.87 10–1 mol e– –

1 mol Fe

2 mol e = 9.35 10–2 mol Fe2+

The change in concentration of Fe2+ will be 2 +29.35 10 mol Fe

0.100 L solution = +0.935 M Fe2+

The final concentration of Fe2+will be:

1.00 M + 0.935 M Fe2+ = 1.94 M Fe2+ The change in concentration of Cu2+ will be

2 +29.35 10 mol Cu

0.100 L solution = –0.935 M Cu2+

The final concentration of Cu+ will be:

1.00 M – 0.935 M Cu2+ = 0.065 M Cu2+

Fe CuSalt Bridge

External circuit

Anode Cathode

Fe2+ (aq) Cu2+ (aq)

(+)(–) electron flow

BIT Chapters 17-20

2+

ocell cell 2

-1 -1

cell -1

FeRTE = E ln

nF Cu

1.94(8.314 J mol K )(298 K) E =0.78 V ln

0.0652(96,500 C mol )

= 0.78 V 0.01284(3.396)

Ecell = 0.736 V

33. (a) 2AgCl(s) + Ni(s) 2Ag(s) + 2Cl–(aq) + Ni2+(aq)

(b) E°cell = E°reduction – E°oxidation = 0.222 V – (–0.257 V) = 0.479 V

34. o

cellE 0.80 V 0.34 V 0.46 V

2+

ocell cell 2

-1 -1

cell -1 2

CuRTE = E ln

nFAg

0.200(8.314 J mol K )(298 K) E =0.46 V ln

2(96,500 C mol ) 0.100

= 0.46 V 0.01284(2.996) V = 0.42 V

A cell potential drop of 10% would be 0.042 V so the new potential would be 0.378 V.

-1 -1

-1 2

2

6.386

2

0.200 x(8.314 J mol K )(298 K)0.378 = 0.46 V ln

2(96,500 C mol ) 0.100 2x

0.200 x 0.082 V = 0.01284 V ln

0.100 2x

0.200 xe 593.65

0.100 2x

x = 0.0399 M

This is the concentration increase that would occur during the time period that the cell potential decreased

by 10 %.

Each cell has a volume of 125 mL.

mol Cu2+ = 0.0399 M x 0.125 L = 4.99 x 10-3 mol Cu2+

BIT Chapters 17-20

t = q/i

3 2

2 –

2 mol e 96,500 C4.99 x 10 mol Cu

mol Cu 1 mol et = 9626 s

0.10 A

or 2.67 hr

For a 125 mL cell, the copper electrode would lose:

63.55 g

0.00499 mol Cu 0.317 g1 mol Cu

The silver electrode would gain:

107.87 g

2 0.00499 mol Ag 1.08 g1 mol Cu

The total mass of the cell remains unchanged according to the law of conservation of mass-energy.