chapter 15 - chemistry - pennsylvania state...
TRANSCRIPT
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 ]
+−−
−= = = = =
If you assume that x << 0.100, then 0.100 - x ≈ 0.100
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+(aq) + H2O(l) HONH2(aq) + H3O+(aq)
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 ]
+−−
−= = = =
If you assume that x << 0.100, then 0.100 - x ≈ 0.100
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+(aq) + H2O(l) HONH2(aq) + H3O+(aq)
HONH3
+ H2O HONH2 H3O+
initial 0.020 M - 0.080 ~0 ∆ -x - +x +x equilibrium 0.020 - x - 0.080 + x x
14
7 3 2a 8 +
3
[H O ][HONH ]1.0 x 10 (x)(0.080 +x)K 9.1 x 10(0.020 -x)1.1 x 10 [HONH ]
+−−
−= = = =
If you assume that x is small:
7 x (0.080 + x) x (0.080) 9.1 x 10
0.020 - x 0.020−≈ =
x = 2.3 x 10-7 M (assumption good)
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+(aq) + H2O(l) HONH2(aq) + H3O+(aq)
HONH3
+ H2O HONH2 H3O+
initial 0.120 M - 0.080 ~0 ∆ -x - +x +x equilibrium 0.120 - x - 0.080 + x x
14
7 3 2a 8 +
3
[H O ][HONH ]1.0 x 10 (x)(0.080 +x)K 9.1 x 10(0.120 -x)1.1 x 10 [HONH ]
+−−
−= = = =
If you assume that x is small:
7 x (0.080 + x) x (0.080) 9.1 x 10
0.120 - x 0.120−≈ =
x = 1.4 x 10-6 M (assumption good)
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+(aq) + H2O(l) HONH2(aq) + H3O+(aq)
HONH3
+ H2O HONH2 H3O+
initial 0.080 M - 0.020 ~0 ∆ -x - +x +x equilibrium 0.080 - x - 0.020 + x x
14
7 3 2a 8 +
3
[H O ][HONH ]1.0 x 10 (x)(0.020 +x)K 9.1 x 10(0.080 -x)1.1 x 10 [HONH ]
+−−
−= = = =
If you assume that x is small:
7 x (0.020 + x) x (0.020) 9.1 x 10
0.080 - x 0.080−≈ =
x = 3.6 x 10-6 M (assumption good)
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+(aq) + H2O(l) HONH2(aq) + H3O+(aq)
HONH3
+ H2O HONH2 H3O+
initial 0.080 M - 0.120 ~0 ∆ -x - +x +x equilibrium 0.080 - x - 0.120 + x x
14
7 3 2a 8 +
3
[H O ][HONH ]1.0 x 10 (x)(0.120 +x)K 9.1 x 10(0.080 -x)1.1 x 10 [HONH ]
+−−
−= = = =
If you assume that x is small:
7 x (0.120 + x) x (0.120) 9.1 x 10
0.080 - x 0.080−≈ =
x = 6.1 x 10-7 M (assumption good)
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
+ −−= = = =
If you assume that x << 0.200, then 0.200 - x ≈ 0.200
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
−≈ =
x = 5.4 x 10-5 M (assumption good)
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(aq) + H2O(l) C2H3O2-(aq) + H3O+(aq)
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)
+ −−= = =
assuming that x is small; 5 x (0.0500 + x) x (0.0500) 1.8 x 100.0500 - x 0.0500
−≈ =
x = 1.8 x 10-5 M (assumption good)
pH = log[H3O+] = 4.74
d. 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 0.1500 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(aq) + H2O(l) C2H3O2-(aq) + H3O+(aq)
HC2H3O2 H2O C2H3O2- H3O+
initial 0.0200 M - 0.0600 ~0 ∆ -x - +x +x equilibrium 0.0200 - x - 0.0600 + x x
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)
+ −−= = =
assuming that x is small; 5 x (0.0600 + x) x (0.0600) 1.8 x 100.0200 - x 0.0200
−≈ =
x = 6.0 x 10-6 M (assumption good)
pH = log[H3O+] = 5.22
d. 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 0.200 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