Question

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

• A. $2 \times 10^{-13} M$
• B. $2 \times 10^{-8} M$
• C. $1 \times 10^{-13} M$
• D. $1 \times 10^{8} M$

Answer 1

The Correct Option is A

To determine the solubility of $Ni(OH)_2$ in 0.1 M NaOH, we must consider the common ion effect and the solubility product constant (K_sp) of $Ni(OH)_2$.

First, the solubility product (K_sp) for $Ni(OH)_2$ is given as $2 \times 10^{-15}$. The dissolution of $Ni(OH)_2$ in water can be represented by the equilibrium reaction:

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

The expression for the solubility product, K_sp, is:

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

Let the solubility of $Ni(OH)_2$ in the presence of 0.1 M NaOH be 's' mol/L. Thus, the concentration of $Ni^{2+}$ ions is s and the concentration of $OH^{-}$ ions will be enhanced by the NaOH solution:

$[OH^{-}] = 0.1 + 2s$

However, since s is very small compared to 0.1, we can approximate $[OH^{-}] \approx 0.1$.

Substitute into the solubility product expression:

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

Solving for 's', we find:

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

Therefore, the solubility of $Ni(OH)_2$ in 0.1 M NaOH is $2 \times 10^{-13} M$.

This matches with the given correct answer option: $2 \times 10^{-13} M$.