Ionic EquilibriumMedium
Question
What is the equilibrium constant of the following reaction? Fe(OH)3(s) + 3H3O+ $\rightleftharpoons$ Fe3+ + 6H2O?
Ksp of Fe(OH)3 = 4 × 10−38
Options
A.2.5 × 10–5
B.4.0 × 104
C.4.0 × 10–4
D.4 × 10–80
Solution
Fe(OH)3(s) $\rightleftharpoons$ Fe3+ (aq) + 3OH– (aq); Ksp = [Fe3+][OH–]3
Now, for given reaction, Fe(OH)3(s) + 3H3O+ $\rightleftharpoons$ Fe3+ + 6H2O
$K_{eq} = \frac{\left\lbrack Fe^{3 +} \right\rbrack}{\left\lbrack H_{3}O^{+} \right\rbrack^{3}} \times \frac{\left\lbrack OH^{-} \right\rbrack^{3}}{\left\lbrack OH^{-} \right\rbrack^{3}} = \frac{K_{sp}}{K_{w}^{3}} $$$= \frac{4 \times 10^{- 38}}{\left( 10^{- 14} \right)^{3}} = 4 \times 10^{4}$$
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