Solutions to Homework Assignment 11
CHM 152
Spring 2002
16.21
Since the acid is monoprotic, the number of moles of KOH is equal to the number of moles of acid.
 1 L   0.08133 mol 
Moles acid = 16.4 mL 

 = 0.00133 mol
1L
 1000 mL  

Molar mass =
16.23
0.2688 g
= 202 g/mol
0.00133 mol
The neutralization reaction is:
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
Since one mole of sulfuric acid combines with two moles of sodium hydroxide, we write:
 0.500 mol H 2SO 4   2 mol NaOH 
mol NaOH = (0.0125 L H2SO4) 
 = 0.0125 mol NaOH

1 L soln

  1 mol H 2SO 4 
concentration of NaOH =
16.24
0.0125 mol NaOH
1000 mL
×
= 0.25 M
50.0 mL soln
1L
2HCOOH + Ba(OH)2 → (HCOO)2Ba + 2H2O
Number of moles of HCOOH reacted =
0.883 mol
× (20.4 × 10−3 L) = 0.0180 mol HCOOH
1L
The mole ratio between Ba(OH)2 and HCOOH is 1:2.
Therefore, the molarity of the Ba(OH)2 solution is:
0.0180 mol HCOOH ×
16.31
16.33
1 mol Ba(OH) 2
1
×
= 0.466 M
2 mol HCOOH
19.3 × 10−3 L
(a)
HCOOH is a weak acid and NaOH is a strong base. Suitable indicators are cresol red and
phenolphthalein.
(b)
HCl is a strong acid and KOH is a strong base. Suitable indicators are all those listed with the
exceptions of thymol blue, bromophenol blue, and methyl orange.
(c)
HNO3 is a strong acid and CH3NH2 is a weak base. Suitable indicators are bromophenol blue, methyl
orange, methyl red, and chlorophenol blue.
The weak acid equilibrium is
+
−
HIn (aq) U H (aq) + In (aq)
We can write a Ka expression for this equilibrium.
Ka =
[H + ][In − ]
[HIn ]
Rearranging,
[HIn ]
[In − ]
[H + ]
Ka
=
+
From the pH, we can calculate the H concentration.
+
[H ] = 10
[HIn]
−
[In ]
−pH
=
= 10
−4
= 1.0 × 10
−4
M
[H + ]
1.0 × 10 −4
=
= 100
Ka
1.0 × 10 −6
−
Since the concentration of HIn is 100 times greater than the concentration of In , the color of the solution
will be that of HIn, the nonionized formed. The color of the solution will be red.
16.35
Solubility is the number of grams of BaSO4 that will dissolve per liter of solution. Molar solubility is the
number of moles of BaSO4 that will dissolve per liter of solution. The solubility product is:
2+
2−
Ksp = [Ba ][SO4 ].
16.36
The Ksp values are very large indicating that there are almost exclusively products in solution at
equilibrium. We assume that soluble ionic compounds completely dissociate into ions. Therefore, the
equilibrium constant is not needed.
16.41
(a)
The solubility equilibrium is given by the equation
−
+
AgI(s) U Ag (aq) + I (aq)
The expression for Ksp is given by
+
−
Ksp = [Ag ][I ]
The value of Ksp can be found in Table 16.2 of the text. If the equilibrium concentration of silver ion
is the value given, the concentration of iodide ion must be
[I − ] =
(b)
Ksp
+
[Ag ]
=
8.3 × 10 −17
91
. × 10
−9
= 9.1 × 10 −9 M
The value of Ksp for aluminum hydroxide can be found in Table 16.2 of the text. The equilibrium
expressions are:
3+
−
Al(OH)3(s) U Al (aq) + 3OH (aq)
3+
− 3
Ksp = [Al ][OH ]
Using the given value of the hydroxide ion concentration, the equilibrium concentration of aluminum
ion is:
Ksp
18
. × 10 −33
= 7.4 × 10 −8 M
[Al 3+ ] =
=
[OH − ]3 ( 2.9 × 10 −9 )3
What is the pH of this solution? Will the aluminum concentration change if the pH is altered?
16.45
The charges of the M and X ions are +3 and −2, respectively (are other values possible?). We first
calculate the number of moles of M2X3 that dissolve in 1.0 L of water
 1 mol 
−19
Moles M 2 X3 = (3.6 × 10−17 g) 
mol
 = 1.3 × 10
288
g


−19
The molar solubility, s, of the compound is therefore 1.3 × 10 M. At equilibrium the concentration of
3+
2−
M must be 2s and that of X must be 3s. (See Table 16.3 of the text.)
2− 3
3+ 2
2
3
5
Ksp = [M ] [X ] = [2s] [3s] = 108s
Since these are equilibrium concentrations, the value of Ksp can be found by simple substitution
5
Ksp = 108s = 108(1.3 × 10
16.46
−19 5
−93
) = 4.0 × 10
Step 1: Write the equilibrium reaction. Then, from the equilibrium equation, write the solubility product
expression.
2+
CaF2 (s) U Ca
2+
−
(aq) + 2 F (aq)
− 2
Ksp = [Ca ][F ]
Step 2: A certain amount of calcium fluoride will dissociate in solution. Let’s represent this amount as −s.
2+
−
2+
−
Since one unit of CaF2 yields one Ca ion and two F ions, at equilibrium [Ca ] is s and [F ] is
2s. We summarize the changes in concentration as follows:
CaF2 (s)
Initial (M):
Change (M):
Equilibrium (M):
U Ca2+ (aq) + 2 F− (aq)
0
+s
s
−s
0
+2s
2s
Recall, that the concentration of a pure solid does not enter into an equilibrium constant expression.
Therefore, the concentration of CaF2 is not important.
2+
Step 3: Substitute the value of Ksp and the concentrations of Ca
product expression to solve for s, the molar solubility.
2+
−
and F in terms of s into the solubility
− 2
Ksp = [Ca ][F ]
4.0 × 10
−11
4.0 × 10
−11
= (s)(2s)
= 4s
2
3
−4
s = molar solubility = 2.2 × 10
The molar solubility indicates that 2.2 × 10
−4
mol/L
mol of CaF2 will dissolve in 1 L of an aqueous solution.
16.48
−
First we can calculate the OH concentration from the pH.
pOH = 14.00 − pH
pOH = 14.00 − 9.68 = 4.32
−
[OH ] = 10
−pOH
−4.32
= 10
= 4.8 × 10
−5
M
The equilibrium equation is:
−
+
MOH (s) U M (aq) + OH (aq)
−
+
From the balanced equation we know that [M ] = [OH ]
−
+
−5 2
−9
Ksp = [M ][OH ] = (4.8 × 10 ) = 2.3 × 10
16.56
(a)
Set up a table to find the equilibrium concentrations in pure water.
PbBr2 (s)
Initial (M)
Change (M)
Equilibrium (M)
U Pb2+ (aq) + 2 Br− (aq)
0
+s
s
−s
2+
0
+2s
2s
− 2
Ksp = [Pb ][Br ]
8.9 × 10
−6
2
= (s)(2s)
s = molar solubility = 0.013 M
(b)
Set up a table to find the equilibrium concentrations in 0.20 M KBr. KBr is a soluble salt that ionizes
−
completely giving a initial concentration of Br = 0.20 M.
PbBr2 (s)
Initial (M)
Change (M)
Equilibrium (M)
U Pb2+ (aq) + 2 Br− (aq)
0
+s
s
−s
2+
0.20
+2s
0.20 + 2s
− 2
Ksp = [Pb ][Br ]
8.9 × 10
−6
8.9 × 10
−6
2
= (s)(0.20 + 2s)
≈ (s)(0.20)
2
−4
s = molar solubility = 2.2 × 10
M
Thus, the molar solubility of PbBr2 is reduced from 0.013 M to 2.2 × 10
−
common ion (Br ) effect.
(c)
−4
M as a result of the
Set up a table to find the equilibrium concentrations in 0.20 M Pb(NO3)2. Pb(NO3)2 is a soluble salt
2+
that dissociates completely giving an initial concentration of [Pb ] = 0.20 M.
PbBr2 (s)
Initial (M):
Change (M):
Equilibrium (M):
−s
U Pb2+ (aq) + 2 Br− (aq)
0.20
+s
0.20 + s
0
+2s
2s
2+
− 2
Ksp = [Pb ][Br ]
8.9 × 10
−6
8.9 × 10
−6
2
= (0.20 + s)(2s)
≈ (0.20)(2s)
2
−3
s = molar solubility = 3.3 × 10
M
Thus, the molar solubility of PbBr2 is reduced from 0.013 M to 3.3 × 10
2+
common ion (Pb ) effect.
−3
M as a result of the
Additional Problem: 10.0 mL of 0.25 M HBr is titrated with 0.10 M KOH. Calculate the pH at the following
points in the titration.
a.
b.
c.
d.
e.
0.00 mL of KOH added
10.0 mL of KOH added
20.0 mL of KOH added
25.0 mL of KOH added
35.0 mL of KOH added
pH = 0.60
pH = 1.12
pH = 1.78
pH = 7.00 (equivalence point)
pH = 12.35