General Chemistry II
Exam 3
Name
SOLUTIONS
1. (10 points) Indicate whether each compound will result in an acidic, basic, or neutral
solution when added to water.
ACIDIC
Fe(NO3)3
BASIC
CH3NH2
NEUTRAL
KBr
ACIDIC
HClO
BASIC
NaHS
ACIDIC
NH4NO3
NEUTRAL
BaCl2
BASIC
NaCN
ACIDIC
KHSO4
BASIC
NaCH3COO
2. (8 points) Indicate the approximate pH range of each of these buffer combinations.
8.4 – 10.4
NaCN/HCN
2.7 – 4.7
formic acid/ potassium formate
3.9 – 5.9
KCH3CH2COO/ CH3CH2COOH
6.2 – 8.2
Na2HPO4/ NaH2PO4
3. (8 points) Use the three buffer combinations listed to answer the questions that
follow.
1
CH3COOH/NaCH3COO
2
H2S/NaHS
3
NH4Cl/NH3
1
Which buffer will be the most acidic?
3
Which buffer would be best to get a pH of 8.4?
2
Which buffer will be essentially neutral?
3
Which of these buffers can not be formed by titrating
a weak acid with a strong base?
4. (10 points) What mass of sodium hypochlorite (NaClO) should be added to 1.0 L of a
0.40 M solution of hypochlorous acid to produce a buffer whose pH is 7.75?
[H O ] = 10
+
− pH
[ClO ] = 1.969
3
−
= 10 −7.75
[HClO]
= 1.778 × 10 −8 M
+
Ka
[ClO ] = 1.969[HClO]
−
[H O ][ClO ]
=
−
= 1.969(0.400)
3
[ClO ] =
[HClO ]
−
[HClO ]
Ka
H 3O +
[
= 0.7876 M
moles NaClO = M ⋅ V
= (0.7876 M )(1.0 L)
]
3.5 × 10 −8
1.778 × 10 −8
= 1.969
=
= 0.7876 mol
0.7876 mol NaClO ×
74.44 g NaClO
= 58.6 g NaClO
1 mol NaClO
5. (12 points) Calculate the pH that results when 32.4 grams of sodium bicarbonate
(NaHCO3) are dissolved in enough water to make 750. mL of solution.
HCO3- + H2O ↔ OH- + H2CO3
[ ]0
0.514
0
0
∆[]
-x
+x
+x
[ ]eq
0.514
x
x
32.4 g NaHCO 3 ×
1 mol NaHCO 3
= 0.3857 mol
84.008 g NaHCO 3
mol
[HCO ] = 0.3857
= 0.514 M
0.750 L
−
3
[H 2 CO 3 ][OH - ]
x2
Kb =
=
= 2.4 × 10 −8
0.514
[HCO 3 ]
x = 1.11 × 10 − 4 = [OH - ]
pOH = −log[OH − ] = 3.954
pH = 14 - pOH = 10.046
6. (6 points) The Ksp of barium carbonate (BaCO3) is 2.6×10-9. How will the solubility
of BaCO3 be altered, compared to its water solubility, (i.e. – increased, decreased, or
unaffected) when it is dissolved in aqueous solutions of these compounds?
UNAFFECTED
NaOH
INCREASED
Na2SO4
UNAFFECTED
CH3OH
DECREASED
BaCl2
INCREASED
HBr
INCREASED
CH3CH2COOH
7. (12 points) The solubility product constant of silver sulfate, Ag2SO4, is 1.2×10-5.
Determine the mass of Ag2SO4 that can be dissolved in 2.00 L of a 0.572-M solution
of sodium sulfate, Na2SO4.
[ ]0
Ag2SO4(s) ↔ 2 Ag+
--0
+
SO40.572
∆[]
-S
+ 2S
+S
[ ]eq
---
2S
0.572
K sp = [Ag + ][Cl − ]
1.2 × 10
−5
moles Ag 2 SO 4 = M ⋅ V
= (2S) (0.6)
S = 0.00229 M
= (0.00229 M )(2.00 L)
= 0.00458 mol
2
0.00458 mol Ag 2 SO 4 ×
311.87 g
= 1.43 g Ag 2 SO 4
1 mole
8. (10 points) A buffer solution is prepared by mixing 350. mL of 0.368 M NaHSO3 and
150. mL of 0.622 M Na2SO3. Determine the pH of the solution.
[NaHSO 3 ] = M 1 V1
=
(0.368 M )(0.350 L)
= 0.2576 M
0.500 L
[Na 2SO 3 ] = M 1 V1
=
(0.622 M )(0.150 L)
= 0.1866 M
0.500 L
V2
V2
HSO3- + H2O ↔ H3O+ + SO3-2
pH = p K a + log
[base]
[acid]
= − log (6.2 × 10 −8 ) + log
= 7.068
(0.1866)
(0.2576)
9. (12 points) Determine the pH that results from mixing 21.6 mL of 0.225 M sodium
hydroxide with 50.0 mL of 0.322 M hypochlorous acid, HClO, whose Ka = 3.5×10-8.
HClO + OH-
ClO- + H2O
→
mol
0.0161
0.00486
0
∆ mol
- 0.00486
-0.00486
+ 0.00486
molfin
[]
VT = 71.6 mL
0.0112
0
0.00486
0.157
0.0679
pH = p K a + log
[base]
[acid]
= − log (3.5 × 10 −8 ) + log
(0.0.0679)
(0.157)
= 7.092
10. (12 points) Pyridine is a weak base whose Kb is 1.7×10-9. A 24.0-mL sample of a
0.180-M solution of pyridine is titrated with 0.140 M hydrochloric acid. Determine
the pH of the titration solution after the addition of 18.5 mL of HCl.
Pyr + H3O+
PyrH+
→
mol
0.00432
0.00259
0
∆ mol
- 0.00259
- 0.00259
+0.00259
molfin
[]
VT = 42.5 mL
0.00173
0
0.00259
0.0407
+ H2O
0.0609
-9
Kb = 1.7×10 ⇒ pKb = 8.770 ⇒ pKa = 5.230
[base]
[acid]
(0.0.407)
= 5.230 + log
(0.0609
= 5.055
pH = p K a + log