TakeHome Assignment #4
The intention of these problems is to provide experience in solving equilibrium problems.
Equilibrium constants for all reactions may be found in the Appendices.
Use ladder
diagrams to help you simplify equilibrium problems.
1.
Calculate the solubility of Zn
2
[Fe(CN)
6
] in (a) distilled water, and (b) in 0.00010 M
K
4
Fe(CN)
6
.
For part (a), let the [Fe(CN)
6
4
] be X.
The [Zn
2+
] is 2X and the K
sp
expression is
(2X)
2
(X) = 4X
3
= 2.1
×
10
16
Solving for X gives a value of 3.7
×
10
6
.
The solubility of Zn
2
[Fe(CN)
6
] is the same
as the concentration of Fe(CN)
6
4
; thus, the molar solubility is X, or 2.1
×
10
16
M.
For part (b) we have two sources of Fe(CN)
6
4
; that coming from the solubility of
Zn
2
[Fe(CN)
6
] and that already present in solution.
Our problem, therefore, is
(2X)
2
(X+ 1.0
×
10
4
) =
2.1
×
10
16
Assuming that X + 1.00
×
10
4
≈
X and solving gives X as 7.2
×
10
7
.
The error in this
assumption is 0.72%, which is small enough to ignore.
The molar solubility is the
amount of Fe(CN)
6
4
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 Fall '11
 Mcdonald
 Analytical Chemistry, Equilibrium, pH, Solubility, mass balance equation, balance equation

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