chapter 15 - chemistry - pennsylvania state...

Chapter 15 : xx, 24, 28, 30, 38, 44, 48, 50, 51, 55, 76, 82, 90, 96(90), 123(111) xx. NH4

+ + OH- NH3 + H2O

NH3 + H3O+ NH4+ + H2O

24. a. HONH2(aq) + H2O(l) OH-(aq) + HONH3

+(aq)

HONH2 H2O OH- HONH3+

initial 0.100 M - ~0 0 ∆ -x - +x +x equilibrium 0.100 - x - x x

2

8 3b

2

[OH ][HONH ] (x)(x) xK 1.1 x 10 [OHNH ] (0.100 -x) 0.100 - x

− +−= = = =

If you assume that x << 0.100, then 0.100 - x ≈ 0.100

2 2

8 x x 1.1 x 100.100 - x 0.100

−≈ =

x2 = 1.1 x 10-9

x = 3.3 x 10-5 M = [OH-] (assumption good)

pOH = log[OH-] = 4.48

pH = 14.00 - pOH = 9.52

b. HONH3Cl(aq) HONH3

+(aq) + Cl-(aq) (Cl- conj. base SA)

HONH3+(aq) + H2O(l) HONH2(aq) + H3O+(aq)

HONH3

+ H2O HONH2 H3O+

initial 0.100 M - 0 ~0 ∆ -x - +x +x equilibrium 0.100 - x - x x

Ka x Kb = Kw for a conjugate acid/base Kb for HONH2 is 1.1 x 10-8.

14 27 3 2

a 8 +3

[H O ][HONH ]1.0 x 10 (x)(x) xK 9.1 x 10(0.100 -x) 0.100 - x1.1 x 10 [HONH ]

+−−

−= = = = =


2 27 x x 9.1 x 10

0.100 - x 0.100−≈ =

x2 = 9.1 x 10-8

x = 3.0 x 10-4 M = [H3O+] (assumption good)

pH = log[H3O+] = 3.52

c. Pure H2O pH = 7.00

d. Using either equilibrium:


HONH3

+ H2O HONH2 H3O+

initial 0.100 M - 0.100 ~0 ∆ -x - +x +x equilibrium 0.100 - x - 0.100 + x x

14

7 3 2a 8 +

3

[H O ][HONH ]1.0 x 10 (x)(0.100 +x)K 9.1 x 10(0.100 -x)1.1 x 10 [HONH ]

+−−

−= = = =


7 x (0.100 + x) x (0.100) 9.1 x 10

0.100 - x 0.100−≈ =

x = 9.1 x 10-7 M (assumption good)

pH = log[H3O+] = 6.04

28. a. The added HCl (as H3O+) reacts with HONH2, forming HONH3

+ (as HONH3Cl): HONH2(aq) + H3O+(aq) HONH3

+(aq) + H2O(l)

HONH2 H3O+ HONH3+ H2O

I 0.100 mole/L x 1.00 L = 0.100 mole

0.020 mole

0 -

F 0.100 - 0.020 = 0.080 mole

0 0.020 mole -

22 2

0.080 mole HONHHONH ] 0.080 M HONH

1.00 L= =[

3

3 30.020 mole HONH

[HONH ] 0.020 M HONH1.00 L

++ += =


HONH3

+ H2O HONH2 H3O+


14

7 3 2a 8 +

3


+−−

−= = = =

If you assume that x is small:

7 x (0.080 + x) x (0.080) 9.1 x 10

0.020 - x 0.020−≈ =


pH = log[H3O+] = 6.64

b. This is a solution of a SA and a WA. The H3O+ from the WA is going to be negligible:

[H3O+] = 0.020 mole/1.00 L = 0.020 M

pH = log[H3O+] = 1.70

c. Same as b. (even more so because H2O is an even weaker acid then HB+).

pH = log[H3O+] = 1.70

d. The added HCl (as H3O+) reacts with just the HONH2, forming HONH3

+ (as HONH3Cl). This adds to the conjugate acid and takes away from the base:

HONH2(aq) + H3O+(aq) HONH3+(aq) + H2O(l)

HONH2 H3O+ HONH3

+ H2O I 0.100 mole/L x 1.00 L =

0.100 mole 0.020 mole

0.100 mole/L x 1.00 L = 0.100 mole

-

F 0.100 - 0.020 = 0.080 mole

0 0.100 + 0.020 = 0.120 mole

-

22 2

0.080 mole HONHHONH ] 0.080 M HONH

1.00 L= =[

3

3 30.120 mole HONH

[HONH ] 0.120 M HONH1.00 L

++ += =


HONH3

+ H2O HONH2 H3O+


14

7 3 2a 8 +

3


+−−

−= = = =


7 x (0.080 + x) x (0.080) 9.1 x 10

0.120 - x 0.120−≈ =


pH = log[H3O+] = 5.86

30. a. This is a solution of a SB and a WB. The OH- from the WB is going to be

negligible:

[OH-] = 0.020 mole/1.00 L = 0.020 M

pOH = log[OH-] = 1.70; pH = 14.00 – pOH = 12.30

b. The added NaOH (as OH-) reacts with HONH3+ (as HONH3Cl), forming HONH2:

HONH3+(aq) + OH-(aq) HONH2(aq) + H2O(l)

HONH3

+ OH- HONH2 H2O

I 0.100 mole/L x 1.00 L = 0.100 mole

0.020 mole

0 -

F 0.100 - 0.020 = 0.080 mole

0 0.020 mole -

2

2 20.020 mole HONH

[HONH ] 0.020 M HONH1.00 L

= =

4

3 30.080 mole NH

[HONH ] 0.080 M HONH1.00 L

++ += =


HONH3

+ H2O HONH2 H3O+


14

7 3 2a 8 +

3


+−−

−= = = =


7 x (0.020 + x) x (0.020) 9.1 x 10

0.080 - x 0.080−≈ =


pH = log[H3O+] = 5.44

c. Same as a. (even more so because H2O is an even weaker base then B).

pOH = log[OH-] = 1.70; pH = 14.00 – pOH = 12.30

d. The added NaOH (as OH-) reacts with HONH3

+ (as HONH3Cl), forming HONH2. This adds to the conjugate base and takes away from the acid:

HONH3+(aq) + OH-(aq) HONH2(aq) + H2O(l)

HONH3

+ OH- HONH2 H2O

I 0.100 mole/L x 1.00 L = 0.100 mole

0.020 mole

0.100 mole/L x 1.00 L = 0.100 mole

-

F 0.100 - 0.020 = 0.080 mole

0 0.100 + 0.020 = 0.120 mole

-

2

2 20.120 mole HONH

[HONH ] 0.120 M HONH1.00 L

= =

4

3 30.080 mole NH

[HONH ] 0.080 M HONH1.00 L

++ += =


HONH3

+ H2O HONH2 H3O+


14

7 3 2a 8 +

3


+−−

−= = = =


7 x (0.120 + x) x (0.120) 9.1 x 10

0.080 - x 0.080−≈ =


pH = log[H3O+] = 6.22

38. [NH3] = 0.75 M

NH4Cl(aq) NH4+(aq) + Cl-(aq)

[NH4+] =

4 44

4 4

1 mole NH Cl 1 mole NH50.0 g NH Cl x x

53.492 g NH Cl 1 mole NH Cl1.00 L

+

= 0.935 M

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

NH3 H2O OH- NH4

+ initial 0.75 M - ~0 0.935 M ∆ -x - +x +x equilibrium 0.75 - x - x 0.935 + x

5 4

b3

[OH ][NH ] (x)(0.935 x)K 1.8 x 10 [NH ] (0.75 -x)

− +− +

= = =

If you assume that x small, then:

-5 x (0.935) 1.8 x 10

0.75=

x = 1.4 x 10-5 M = [OH-] (assumption good)

pOH = log[OH-] = 4.84

pH = 14.00 – pOH = 9.16

44. a. + pH 7.40 8

3[H O ] = 10 = 10 = 3.98 x 10 M− − −

+ 73a 3

2 3

[HCO ]K = [H O ] = 4.3 x 10

[H CO ]

−−

7 7

3+ 8

2 3 3

[HCO ] 4.3 x 10 4.3 x 10 = = [H CO ] [H O ] 3.98 x 10

− − −

− = 10.8

2 3

3

CO ] 1 = 10.8[HCO ]−

[H = 0.093

b. [ + pH 7.15 8

3H O ] = 10 = 10 = 7.08 x 10 M− − −

2

+ 84a 3

2 4

[HPO ]K = [H O ] = 6.2 x 10

[H PO ]

−−

−

2 8 8

4+ 8

2 4 3

[HPO ] 6.2 x 10 6.2 x 10 = = [H PO ] [H O ] 7.08 x 10

− − −

− − = 0.876

2 42

4

PO ] 1 = 0.876[HPO ]

−

−

[H = 1.1

c. The best buffer is one with a 1:1 ratio of the acid (or base) to the conjugate base (or conjugate acid). This will give a pH close to pKa. The pKa of H3PO4 is 2.12, which is far removed from 7.1 – 7.2. There would be very little H3PO4 at that basic a pH.

48. a. SA + CB of SA not a buffer – an acidic solution b. SA + WA not a buffer – no conjugate base produced. Acidic.

c. SA + CB of WA yes, a buffer. Half of the F- ion would react with the H3O+ from HNO3 to give HF. This results in a mixture of HF (a WA) and F- (its conjugate base) – a buffer.

d. SA + SB not a buffer – a basic solution (NaOH in excess). 50. Henderson – Hasselbalch is convenient for this type of calculation.

a[A ]pH = pK + log[HA]

−

The reaction is:

C2H3O2

-(aq) + HCl(g) HC2H3O2(aq) + Cl-(aq) a. For pH to equal pKa, the ratio must be 1:1. There are

2 3 22 3 2

mole C H O x 1.0 L = 1.0 mole C H O

L

−1.0 −

To convert half (0.50 mole) into would require 0.50 mole HCl. 2 3 2HC H O

b. [A ]og[HA]

−

4.00 = 4.74 + l

[A ]log = 0.54[HA]

− −

;

[A ]log[HA] 0.54[A ] = = 10 = 0.288

[HA]

− − − 10

Since x moles of are produced from x moles of HCl(g) (by removing x moles ), the equation above becomes:

2 3 2HC H O

2 3 2C H O −

1.0 = 0.288x

x− ; x = 0.78 = mole HCl(g)

c. [A ]5.00 = 4.74 + log[HA]

−

[A ]og = 0.26[HA]

−

l ; [A ]log[HA] 0.26[A ] = = 10 = 1.82

[HA]

− −

10

1.0 = 1.82x

x− ; x = 0.35 = mole HCl(g)

51.

55. a. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2

-(aq)

HC2H3O2 H2O H3O+ C2H3O2-

initial 0.200 M - ~0 0 ∆ -x - +x +x equilibrium 0.200 - x - x x

2

5 3 2 3 2a

2 3 2

[H O ][C H O ] (x)(x) xK 1.8 x 10 [HC H O ] (0.200 -x) 0.200 - x

+ −−= = = =


2 2

-5 x x 1.8 x 100.200 - x 0.200

≈ =

x2 = 3.6 x 10-6

x = 1.9 x 10-3 M = [H3O+] (assumption good)

pH = log[H3O+] = 2.72

b. The added KOH (as OH-) reacts with HC2H3O2, forming C2H3O2-:

HC2H3O2(aq) + OH-(aq) C2H3O2

-(aq) + H2O(l)

HC2H3O2 OH- C2H3O2-

I 0.200 mole/L x 0.100 L = 0.0200 mole

0.100 mole/L x 0.0500 L = 0.00500 mole

0

F 0.0200 - 0.00500 = 0.0150 mole

0 0.0050 mole

2 3 2

2 3 2 2 3 20.0150 mole HC H O

[HC H O ] 0.100 M HC H O0.150 L

= =

2 3 22 3 2 2 3 2

0.00500 mole C H O[C H O ] 0.0333 M C H O

0.150 L

−− −= =

HC2H3O2(aq) + H2O(l) C2H3O2-(aq) + H3O+(aq)

HC2H3O2 H2O C2H3O2

- H3O+ initial 0.100 M - 0.0333 ~0 ∆ -x - +x +x equilibrium 0.100 - x - 0.0333 + x x

5 3 2 3 2

a2 3 2

[H O ][C H O ] (x)(0.0333 +x)K 1.8 x 10 [HC H O ] (0.100 -x)

+ −−= = =

assuming that x is small; 5 x (0.0333 + x) x (0.0333) 1.8 x 100.100 - x 0.100

−≈ =


pH = log[H3O+] = 4.27

c. The added KOH (as OH-) reacts with HC2H3O2, forming C2H3O2

-: HC2H3O2(aq) + OH-(aq) C2H3O2

-(aq) + H2O(l)

HC2H3O2 OH- C2H3O2-

I 0.200 mole/L x 0.100 L = 0.0200 mole

0.100 mole/L x 0.100 L = 0.0100 mole

0

F 0.0200 - 0.0100 = 0.0100 mole

0 0.0100 mole

2 3 22 3 2 2 3 2

0.0100 mole HC H O[HC H O ] 0.0500 M HC H O

0.200 L= =

2 3 22 3 2 2 3 2

0.0100 mole C H O[C H O ] 0.0500 M C H O

0.200 L

−− −= =


HC2H3O2 H2O C2H3O2

- H3O+ initial 0.0500 M - 0.0500 ~0 ∆ -x - +x +x equilibrium 0.0500 - x - 0.0500 + x x

5 3 2 3 2

a2 3 2

[H O ][C H O ] (x)(0.0500 +x)K 1.8 x 10 [HC H O ] (0.0500 -x)

+ −−= = =


−≈ =


pH = log[H3O+] = 4.74

d. The added KOH (as OH-) reacts with HC2H3O2, forming C2H3O2


-(aq) + H2O(l)

HC2H3O2 OH- C2H3O2-

I 0.200 mole/L x 0.100 L = 0.0200 mole

0.100 mole/L x 0.1500 L = 0.0150 mole

0

F 0.0200 - 0.0150 = 0.00500 mole

0 0.0150 mole

2 3 22 3 2 2 3 2

0.0050 mole HC H O[HC H O ] 0.0200 M HC H O

0.250 L= =

2 3 2

2 3 2 2 3 20.01500 mole C H O

[C H O ] 0.0600 M C H O0.250 L

−− −= =


HC2H3O2 H2O C2H3O2- H3O+


5 3 2 3 2

a2 3 2

[H O ][C H O ] (x)(0.0600 +x)K 1.8 x 10 [HC H O ] (0.0200 -x)

+ −−= = =


−≈ =


pH = log[H3O+] = 5.22

d. The added KOH (as OH-) reacts with HC2H3O2, forming C2H3O2


-(aq) + H2O(l)

HC2H3O2 OH- C2H3O2-

I 0.200 mole/L x 0.100 L = 0.0200 mole

0.100 mole/L x 0.200 L = 0.0200 mole

0

F 0.0200 - 0.0200 = 0.0 mole

0 0.0200 mole

2 3 2

2 3 2 2 3 20.0200 mole C H O

[C H O ] 0.0667 M C H O0.300 L

−− −= =

C2H3O2

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

C2H3O2- H2O HC2H3O2 OH-

initial 0.0667 M - 0 ~0 ∆ -x - +x +x equilibrium 0.0667 - x - x x

2

10w 2 3 2b

a 2 3 2

K [OH ][HC H O ] xK = 5.6 x 10 K (0.0667 -x)[C H O ]

−−

−= = =

assuming that x is small; 2 2

10 x x 5.6 x 100.0667 - x 0.0667

−≈ =

x2 = 3.7 x 10-11; x = 6.1 x 10-6 M (assumption good)

pOH = log[OH-] = 5.21; pH = 14.00 – pOH = 8.79

f. This is 0.100 mole/L x 0.250 L = 0.0250 mole OH- excess 0.0250 mole OH- – 0.0200 mole OH- reacted = 0.00500 mole OH-

0.00500 mole OHOH ] 0.0143 M OH0.350 L

−− −= =[

pOH = log[OH-] = 1.85; pH = 14.00 – pOH = 12.15

76. a. Ag2CO3(s) 2 Ag+(aq) + CO3

2-(aq) b. Ce(IO3)3(s) Ce3+(aq) + 3 IO3

-(aq) c. BaF2(s) Ba2+(aq) + 2 F-(aq) 82. a. PbI2(s) Pb2+(aq) + 2 I-(aq)

PbI2 Pb2+ I- initial - 0 0 ∆ - +x +2x equilibrium - x 2x

Ksp = [Pb2+][I-]2 = 1.4 x 10-8 = x(2x)2 = 4x3 x = 1.5 x 10-3 M = [Pb2+] = [PbI2] = molar solubility

b. CdCO3(s) Cd2+(aq) + CO3

2-(aq)

CdCO3 Cd2+ CO32-

initial - 0 0 ∆ - +x +x equilibrium - x x

Ksp = [Cd 2+][CO3

2-] = 5.2 x 10-12 = x2 x = 2.3 x 10-6 M = [Cd2+] = [CO3

2-] = [CdCO3] = molar solubility c. Sr3(PO4)2(s) 3 Sr 2+(aq) + 2 PO4

3-(aq)

Sr3(PO4)2 Sr 2+ PO43-

initial - 0 0 ∆ - +3x +2x equilibrium - 3x 2x

Ksp = [Sr2+]3[PO4

3-]2 = 1 x 10-31 = (3x)3(2x)2 = 108x5 x = 2 x 10-7 M = [Sr3(PO4)2] = molar solubility

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

Ag2SO4 Ag+ SO4

2- initial - 0.10 M 0 ∆ - +2x +x equilibrium - 0.10 + 2x x

Ksp = [Ag+]2[SO4

2-] = 1.2 x 10-5 = (2x)2x = 4x3

x = 1.4 x 10-2 M = [SO4

2-] = molar solubility

b. AgNO3(aq) Ag+(aq) + NO3-(aq) (ionizes completely)

[Ag+] = [AgNO3] = 0.10 M

Ag2SO4(s) 2 Ag+(aq) + SO42-(aq)

Ag2SO4 Ag+ SO4

2- initial - 0.10 M 0 ∆ - +2x +x equilibrium - 0.10 + 2x x

Ksp = [Ag+]2[SO4

2-] = 1.2 x 10-5 = (0.10 + 2x)2x

Assume x small: 1.2 x 10-5 ≈ (0.10)2x

x = 1.2 x 10-3 M (assumption good) = [SO42-] = molar solubility

c. K2SO4(aq) 2 K+(aq) + SO4

2-(aq) (ionizes completely) [SO4

2-] = [K2SO4] = 0.20 M

Ag2SO4(s) 2 Ag+(aq) + SO42-(aq)

Ag2SO4 Ag+ SO4

2- initial - 0 0.20 M ∆ - +2x +x equilibrium - 2x 0.20 + x

Ksp = [Ag+]2[SO4

2-] = 1.2 x 10-5 = (2x)2(0.20 + x)

Assume x small: 1.2 x 10-5 ≈ 4x2(0.20)

x = 3.9 x 10-3 M (assumption good) = [SO42-] = molar solubility

90. a. AgF, because F- is the CB of a WA b. Pb(OH)2, because OH- is a SB c. Sr(NO2)2, because NO2

- is the CB of a WA d. Ni(CN)2, because CN- is the CB of a WA 111. a. HC2H3O2 + OH- C2H3O2

- + H2O K1 = ?

The reverse is Kb of the CB of a WA:

C2H3O2- + H2O HC2H3O2 + OH-

14

wb 5

a

K 1.0 x 10K = =K 1.8 x 10

−

−= 5.6 x 10-10

1 10b

1 1K = =K 5.6 x 10−= 1.8 x 109

b. C2H3O2

- + H3O+ HC2H3O2 + H2O K1 = ?

The reverse is Ka of a WA:

HC2H3O2 + H2O C2H3O2- + H3O+

1 5a

1 1K = =K 1.8 x 10−= 5.6 x 104

c. This reaction is just: H3O+ + OH- 2 H2O K1 = ?

This is the reverse of the Kw reaction: 2 H2O H3O+ + OH- Kw = 1.0 x 10-14

1 14w

1 1 =K 1.0 x 10−=K = 1.0 x 1014

chapter 15 - chemistry - pennsylvania state...

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