Determine the pH and pOH of the following 4 solutions:
a) A 4.5 x 10-3 M HBr solution.
Strong acid: pH = -log (H3O+) = 2.35, pOH = 14 – 2.35 = 11.65
b)
A 3.67 x 10-5 M KOH solution.
Strong base: pOH = -log (OH-) = 4.435; pH = 14 – 4.435 = 9.565
c) A solution made by diluting 25 mL of 6.0 M HCl until the final volume of the solution is
1.75 L.
M1V1 = M2V2
(6.0M)(0.025 L) = (M2)(1.75 L)
M2 = 0.0857 M; pH = 1.07; pOH
= 12.93
d) 5.0 L of an aqueous solution that contains 1.0 grams of HBr and 1.0 grams of nitric acid.
1.0 g HBr (1 mol / 79.908 g) = 0.0125 mol;
1.0 g HNO3 (1 mol / 63.008) = 0.0159 mol
total moles = 0.0284 mol / 5.0 L = 0.00568 M; since both acids are strong, pH = -log (conc.)
pH = 2.25; pOH = 11.75
Identify the following as strong or weak acids, strong or weak bases, neutral salts, basic salts or
acidic salts.
LiCl __ neutral salt _______, NH3_______ weak base____, NH4ClO4___ acidic salts _____,
HClO4__ strong acid__,
Ba(OH)2__ strong base ___, Ba(CH3COO)2__basic salts __,
HF__weak acid ______,
NaF_____ basic salt_____,
KOH_____ strong base ____,
AlBr3____ acidic salt __,
K2CO3___ basic salt______,
Ca(NO3)2___ neutral salts ____.
Write a balanced chemical equation to show that sodium cyanide (NaCN) is basic in water
solution.
NaCN  Na+ + CN-
CN- + H2O  HCN + OH-
57. What is the Kb expression and value for the equilibrium above?
Kb = [HCN][OH-] / [CN-]
Kb = Kw / Ka = 1.0 x 10-14 / 4.9 x 10-10 = 2.0 x 10-5
58. Which one of the following salts, when dissolved in water, produces the solution with the
highest pH?
a. KI
b. KBr
c. KF
d. KCl
If an equal number of moles of HCN and KOH are added to water, what is the resulting solution:
acidic, basic, or neutral? Explain.
KOH is a strong base; HCN is a weak acid. When reacted together, they’ll produce water and
KCN. This salt will dissolve 100%, leaving K+ (no further reaction) and CN- (which will react
with water). This will produce a basic solution (see eqns from #56).
60. Calculate the pH of a 0.100 M NaCH3COO solution. Ka for acetic acid is 1.8 x 10-5.
(Use A- as CH3COO-):
A- + H2O  HA + OH0.100
0
0
Kb = x2 / (0.100 – x) = 5.6 x 10-10
-x
+x
+x
Assume x << 0.100 (and check it!)
0.100 – x
x
x
Kb = x2 / 0.100 = 5.6 x 10-10
x2 = 5.6 x 10-11; x = 7.48 x 10-6; pOH = 5.13, pH =
8.87
61. Calculate the pH of a 0.100 M CH3NH3Cl solution. Kb for methylamine, CH3NH2, is 3.7 x
10-4.
(Use BH+ as CH3NH3+):
BH+ + H2O  H3O+ + B
0.100
0
0
Ka = x2 / (0.100 – x) = 2.7 x 10-11
-x
+x
+x
Assume x << 0.100 (and check it!)
0.100 – x
x
x
Ka = x2 / 0.100 = 2.7 x 10-11
x2 = 2.7 x 10-12; x = 1.64 x 10-6; pH = 5.78
A 50.0 mL sample of 0.500 M HC2H3O2 acid is titrated with 0.150 M NaOH. Ka = 1.8x10-5 for
HC2H3O2. Calculate the pH of the solution after the following volumes of NaOH have been
added: a) 0 mL; b) 166.7 mL; c) 180.0 mL.
a) 0 ml of base; only a weak acid is initially present so [H+] ≠ [HA]
HC2H3O2
I
C
E
0.500
-x
0.50-x
H+ + C2H3O20
x
x
0
x
x

Ka =
[ H  ][C 2 H 3 O2 ]
[ HC2 H 3 O2 ]
1.8x10-5 =
[H +] = x = 0.500 ( 1.8x10 5 ) = 3.0x10-3
x2
0.500
pH = -log 3.0x10-3 = 2.52
b) 166.7 ml of NaOH are added

1 L  0.500 moles HC 2 H 3 O2

L
 1000 mL 
moles HC2H3O2 = 50 .0 ml 

1 L  0.150 moles NaOH 


L

 1000 mL 
moles NaOH = 166 .7 ml 

 

2.50x10-2 moles HC2H3O2
2.50x10-2 moles NaOH
neutralization: HC2H3O2 + OH-  C2H3O2- + H2O
I
C
Final
0.0250
-0.0250
0
0.0250
-0.0250
0
0
+0.0250
0.0250
only acetate remains – a weak base:
[C2H3O2-] =
2.50 10 2 moles
 0.115 M
0.2167 L
base hydrolysis: C2H3O2- + H2O
I
C
E
Kb for C2H3O2- =
HC2H3O2 + OH-
0.115
-x
0.115-x
0
x
x
1  10 14
= 5.6x10-10
1.8 x10 5
Kb =
0
x
x
[ HC 2 H 3O2 ][OH  ]
[C 2 H 3O2 ]
5.6x10-10 =
x2
0.115
x = [OH-] =


0.115 5.6  10 10 = 8.0x10-6
pOH = -log 8.0x10-6 = 5.10
pH = 14 – 5.10 = 8.90
 At the equivalence point for a WA/SB titration, the pH > 7 due to the OHproduced by the conjugate base hydrolysis reaction.
c) 180.0 mL of NaOH are added
from part b, moles HC2H3O2 = 2.50x10-2 moles HC2H3O2
 1 L  0.150 moles NaOH 

moles NaOH = 180.00 ml
  2.70x10-2 moles NaOH
L

 1000 mL 
moles excess base = 2.70x10-2 moles - 2.50x10-2 moles = 2.0x10-3 moles NaOH
M OH- = M NaOH =
pOH = -log 8.7x10-3 = 2.06
2.0 x10 3 moles
 8.7x10-3 M OH0.2300 L
pH = 14 – 2.06 = 11.94
*Excess NaOH remains - this is the primary source of OH-. We can neglect the
hydrolysis of the conjugate base because this would contribute a relatively small
amount of OH- compared to the amount that comes directly from the excess NaOH.
What is the concentration of a saturated silver acetate solution? Ksp(AgC2H3O2) = 1.94 x 10-3.
25.
Since Ksp = [Ag+][C2H3O2-], and the concentration of silver ions is the same as the
concentration of acetate ions, we can set up the following equation:
1.94 x 10-3 = x2
x = 0.0440 M
What is the concentration of a saturated lead chloride solution? Ksp(PbCl2) = 1.17 x 10-5.
Ksp = [Pb+2][Cl-]2. Since the concentration of chloride ions is twice that of lead (II)
ions, this boils down to the following equation:
1.17 x 10-5 = (x)(2x)2
1.17 x 10-5 = 4x3
Silver phosphate, Ag3PO4, is an insoluble salt that has a Ksp = 1.3 x 10-20.
a) Calculate the molar solubility of Ag3PO4 in pure water.
Ag3PO4(s)
I
C
E
3Ag+(aq) + PO43-(aq)
0
3x
3x
Ksp = [Ag+]3[PO43-]
0
x
x
Ksp = (3x)3x
1.3x10-20 = 27x4
x4 = 4.8x10-22
x = 4.7x10-6 M = molar solubility of Ag3PO4 in pure water
b) Calculate the molar solubility of Ag3PO4 in a solution containing 0.020 M Na3PO4 (a
soluble salt).
soluble salt: Na3PO4  3Na+ + PO43Phosphate is the common ion:
[PO43-] = [Na3PO4] = 0.020 M (since 1 mol Na3PO4 forms 1 mol PO43- ions)
Ag3PO4(s)
3Ag+(aq) + PO43-(aq)
I
C
E
0
3x
3x
0.020
x
0.020+x
Ksp = [Ag+]3[PO43-]
1.3x10-20 = = (3x)30.020
6.5x10-19 = 27x3
x3 = 2.4x10-20
x = 2.9x10-7M = molar solubility of Ag3PO4 with a common ion
 Adding common ion decreases the solubility of Ag3PO4
17. a. Calculate the standard free energy change, ΔG, for the following at 25 C:
MgO(s) + C(graphite)  Mg(s) + CO(g)
ΔH = 491.18 kJ
ΔS = 197.67 J/K
G = H - TS
J   1 kJ 

 = + 432.27 kJ
G = 491.18 kJ - (298 K) 197.67  
K   1000 J 

b. Is this reaction spontaneous at 25 C? If not, at what temperature can we make this reaction
spontaneous?
No, it is not spontaneous at 25 C (G is a large positive value)
Set G = 0:
T=
0 = H - TS
491.18 kJ
= 2484.8 K
J  1 kJ 


197.67 
K  1000 kJ 


T=
H
S
 This is T at equilibrium, spont at T >
2484.8 K
18. For the unbalanced reaction 2 SO2(g) + O2(g) → 2 SO3(g) calculate ΔG at 25.0ºC
when the reactants and product are at the following partial pressures: 10.0 atm SO2 , 10.0
atm O2 , and 1.00 atm SO3.
G = Go + RT ln Q
Q = 12 / (102 * 10) = 1/1000
Go = [ 2mol(-371.1 kJ/mol)] – [ 2mol(-300.2 kJ/mol) + 0 ] = -141.8 kJ
G = -141.8 kJ + (8.314 J/mol·K) (298 K) ( kJ/1000J) (ln 1/1000) = -159 kJ
In each of the following equations, indicate the element that has been oxidized and the one that
has been reduced. You should also label the oxidation state of each before and after the process:


2 Na + FeCl2  2 NaCl + Fe
Sodium is oxidized, going from a 0 to +1 oxidation state.
Iron is reduced, going from a +2 to 0 oxidation state.


2 C2H2 + 5 O2  4 CO2 + 2 H2O
Carbon is oxidized, going from a –1 to +4 oxidation state.
Oxygen is reduced, going from a 0 to –2 oxidation state.

2 PbS + 3 O2  2 SO2 + 2 PbO
Sulfur is oxidized, going from a –2 to +4 oxidation state.
1.
2.
3.

11.
Oxygen is reduced, going from a 0 to –2 oxidation state.
A galvanic cell is constructed based on the following reactions:
Zn2+ + 2e-  Zn
E red = -0.76 V
Al3+ + 3e-  Al
E red = -1.66 V
a) Write the overall balanced equation for the cell and calculate the cell emf under standard
conditions.
3(Zn2+ + 2e-  Zn)
E red = -0.76 V
2(Al  Al3+ + 3e-)
E ox = +1.66 V
3Zn2+ + 2Al  3Zn + 2Al3+
Eocell= -0.76 V + 1.66 V = 0.90 V
b) Calculate G° at 298 K.
 96500 J 
 0.90 V = -5.2 x105 J or -520 kJ
G = -nFE = -(6 mol e-) 
 V  mol e  
c) What is the value of the equilibrium constant, K, for this reaction at 298 K?
Eo = (RT/nF) ln K; n = 6; K = 1.6 x 1091
Which combination below will undergo a spontaneous reaction?
Half Reaction Table
E red
–
Br2(l) + 2e-  2Br (aq)
+1.06 V
2+
Cu (aq) + 2e-  Cu(s)
+0.34 V
Ni2+(aq) + 2e-  Ni(s)
−0.28 V
3+
Al (aq) + 3e-  Al(s)
−1.66 V
a.
b.
c.
d.
e.
Ni 2+(aq) with Br–(aq)
Cu(s) with Al3+(aq)
Ni (s) with Br–(aq)
Br–(aq) with Cu2+(aq)
Br2(l) with Ni(s)
A voltaic cell is constructed that used the following reaction and operates at 298 K:
2Al(s) + 3Mn2+(aq)  2Al3+(aq) + 3Mn(s)
a. What is the EMF of this cell under standard conditions? (Use Appendix D for Eored
values.)
+1.66 V + -1.18 V = 0.48 V
b. What is the EMF of this cell when [Mn2+] = 0.10 M and [Al3+] = 1.5 M?
E = 0.48 V – (0.0592 V / 6 mol e-) (log (1.5)2/(0.10)3) = 0.45 V