First slide

Ionic equilibrium

Question

What is the molar solubility of $Fe {(OH)}_{2} (K_{sp} = 8.0 \times 10^{- 16})$ is $pH = 13.0 ?$

Moderate

A

$8.0 \times 10^{- 18}$
B

$8.0 \times 10^{- 15}$
C

$8.0 \times 10^{- 17}$
D

$8.0 \times 10^{- 14}$

Solution

$\begin{array}{l} Fe (OH)_{2} \to {Fe}^{2 +} + 2 {OH}^{-} \\ P^{H} + P^{OH} = 14 \\ P^{H} = 13 \Rightarrow P^{OH} = 1 \Rightarrow [{OH}^{-}] = 10^{- 1} \\ \Rightarrow 2 s = [{OH}^{-}] = 10^{- 1} \\ K_{Sp} = ({Fe}^{+ 2}) {({OH}^{-})}^{2} \\ \Rightarrow 8 \times 10^{- 16} = (S) {(10^{- 1)})}^{2} \\ \Rightarrow S = 8 \times 10^{- 14} \end{array}$

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