1667 the solubility product expression for magnesium

16.67 The solubility product expression for magnesium hydroxide is
K
sp
= [Mg
2 +
][OH
−
]
2
= 1.2 × 10
− 11
We find the hydroxide ion concentration when [Mg
2 +
] is 1.0 × 10
− 10
M
1
11
2
10
1.2 10
1.0 10
−
−
⎛ ⎞
×
= = ⎜ ⎟
⎜ ⎟
×
⎝ ⎠
[OH ] 0.35
−
M
Therefore the concentration of OH
−
must be slightly greater than 0.35 M .
16.68 We first determine the effect of the added ammonia. Let's calculate the concentration of NH
3
. This is a
dilution problem.
M
i
V
i
= M
f
V
f
(0.60 M )(2.00 mL) = M
f
(1002 mL)
M
f
= 0.0012 M NH
3
Ammonia is a weak base ( K
b
= 1.8 × 10
− 5
).
NH
3
+ H
2
O
U
NH
4
+
+ OH
−
Initial ( M ): 0.0012 0 0
Change ( M ): − x + x + x
Equil. ( M ): 0.0012 − x x x
4
b
3
[NH ][OH ]
[NH ]
+ −
= K

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 482
2
5
1.8 10
(0.0012 )
−
× =
−
x
x
Solving the resulting quadratic equation gives x = 0.00014, or [OH
−
] = 0.00014 M
This is a solution of iron(II) sulfate, which contains Fe
2 +
ions. These Fe
2 +
ions could combine with OH
−
to
precipitate Fe(OH)
2
. Therefore, we must use K
sp
for iron(II) hydroxide. We compute the value of Q
c
for
this solution.
Fe(OH)
2
( s )
U
Fe
2 +
( aq ) + 2OH
−
( aq )
Q = [Fe
2 +
]
0
[OH
−
]
0
2
= (1.0 × 10
− 3
)(0.00014)
2
= 2.0 × 10
− 11
Note that when adding 2.00 mL of NH
3
to 1.0 L of FeSO
4
, the concentration of FeSO
4
will decrease slightly.
However, rounding off to 2 significant figures, the concentration of 1.0 × 10
− 3
M does not change. Q is larger
than K
sp
[Fe(OH)
2
] = 1.6 × 10
− 14
. The concentrations of the ions in solution are greater than the equilibrium
concentrations; the solution is saturated. The system will shift left to reestablish equilibrium; therefore, a
precipitate of Fe(OH)
2
will form .
16.71 First find the molarity of the copper(II) ion
4
1 mol
Moles CuSO 2.50 g 0.0157 mol
159.6 g
= × =
2
0.0157 mol
[Cu ] 0.0174
0.90 L
+
= = M
As in Example 16.15 of the text, the position of equilibrium will be far to the right. We assume essentially all
the copper ion is complexed with NH
3
. The NH
3
consumed is 4 × 0.0174 M = 0.0696 M . The uncombined
NH
3
remaining is (0.30 − 0.0696) M , or 0.23 M . The equilibrium concentrations of Cu(NH
3
)
4
2 +
and NH
3
are
therefore 0.0174 M and 0.23 M , respectively. We find [Cu
2 +
] from the formation constant expression.
2
13 3 4
f
2 4 2 4
3
[Cu(NH ) ]
0.0174
5.0 10
[Cu ][NH ] [Cu ](0.23)
+
+ +
= = × = K
[Cu
2 +
] = 1.2 × 10
− 13
M
16.72 Strategy: The addition of Cd(NO
3
)
2
to the NaCN solution results in complex ion formation. In solution,
Cd
2 +
ions will complex with CN
−
ions. The concentration of Cd
2 +
will be determined by the following
equilibrium
Cd
2 +
( aq ) + 4CN
−
( aq )
U
Cd(CN)
4
2 −
From Table 16.4 of the text, we see that the formation constant ( K
f
) for this reaction is very large
( K
f
= 7.1 × 10
16
). Because K
f
is so large, the reaction lies mostly to the right. At equilibrium, the
concentration of Cd
2 +
will be very small. As a good approximation, we can assume that essentially all the
dissolved Cd
2 +
ions end up as Cd(CN)
4
2 −
ions. What is the initial concentration of Cd
2 +
ions? A very small
amount of Cd
2 +
will be present at equilibrium. Set up the K
f
expression for the above equilibrium to solve
for [Cd
2 +
].

CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 483
Solution: Calculate the initial concentration of Cd
2 +
ions.