Kelvi

Kelvi

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Kelvi

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Find out the solubility of Ni(OH) $_2$ in 0.1 M NaOH. Given that the ionic product of Ni(OH) $_2$ is $2 \times 10^{-15}.$

medium

Equilibrium

2020

chemistry

$2 \times 10^{-8}$ M

$1 \times 10^{-13}$ M

$1 \times 10^{8}$

Explanation

To find the solubility of $\mathrm{Ni(OH)_2}$ in 0.1 M NaOH, we need to consider the common ion effect. $\\$ Given: $\\$ • Ionic product $(K_{sp})$

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M

$2 \times 10^{-13}$ M

of

\mathrm{Ni(OH)_2} = 2 \times 10^{-15} \\

• Concentration of

\mathrm{OH^-}

from NaOH is 0.1 M.

\\

The dissolution of

\mathrm{Ni(OH)_2}

can be represented as:

\\ \mathrm{Ni(OH)_2 (s) \rightleftharpoons Ni^{2+} (aq) + 2OH^- (aq)} \\

Let

s

be the solubility of

\mathrm{Ni(OH)_2}

in mol/L.

\\

At equilibrium, the concentration of

\mathrm{Ni^{2+}}

will be

s

and the concentration of

\mathrm{OH^-}

will be

0.1 + 2s. \\

Since

s

is very small compared to 0.1, we can approximate

0.1 + 2s \approx 0.1. \\

The expression for the ionic product is:

\\ K_{sp} = [\mathrm{Ni^{2+}}][\mathrm{OH^-}]^2 \\

Substitute the equilibrium concentrations:

\\ 2 \times 10^{-15} = s \cdot (0.1)^2 \\

Simplify:

\\ 2 \times 10^{-15} = s \cdot 0.01 \\ s = \frac{2 \times 10^{-15}}{0.01} \\ s = 2 \times 10^{-13} \\

Therefore, the solubility of

\mathrm{Ni(OH)_2}

in 0.1 M NaOH is

2 \times 10^{-13}

M.

\\

This corresponds to Option 4.