Question

Solubility of $\mathrm{Cu}{\left(\mathrm{OH}\right)}_2$ in $0.1\mathrm{M}$ aqueous $\mathrm{NaOH}$ solution is ($\mathrm{Ksp}$ of $\mathrm{Cu}{\left(\mathrm{OH}\right)}_2=2.2\times {10}^{-20}$)

Answer 1

Calculating the solubility of $\mathrm{Cu}{\left(\mathrm{OH}\right)}_2$

Step 1: Analyzing the given data

Concentration of $\mathrm{NaOH}=0.1\mathrm{M}$

$\mathrm{Ksp}$ of $\mathrm{Cu}{\left(\mathrm{OH}\right)}_2=2.2\times {10}^{-20}$

Step 2: Calculating solubility of $\mathrm{Cu}{\left(\mathrm{OH}\right)}_2$

Sodium hydroxide is a strong electrolyte that completely dissociates into its constituents ions in aqueous solution.

$\underset{\mathrm{Sodium}\mathrm{hydroxide}}{\mathrm{NaOH}\left(\mathrm{l}\right)}\to \underset{(0.1\mathrm{M})}{\underset{\mathrm{Sodium}}{{\mathrm{Na}}^{+}}}+\underset{(0.1\mathrm{M})}{\underset{\mathrm{Hydroxide}}{{\mathrm{OH}}^{-}}}$

Let the solubility of $\mathrm{Cu}{\left(\mathrm{OH}\right)}_2$ be “s”

$\underset{\mathrm{Copper}\mathrm{oxide}}{\mathrm{Cu}{\left(\mathrm{OH}\right)}_2\left(\mathrm{l}\right)}\leftrightharpoons \underset{\left(\mathrm{s}\right)}{\underset{\mathrm{Copper}}{{\mathrm{Cu}}^{2+}}}+\underset{(0.1+2\mathrm{s})}{\underset{\mathrm{Hydroxide}}{2{\mathrm{OH}}^{-}}}$

$\mathrm{Ksp}=\left[{\mathrm{Cu}}^{+2}\right]{\left[{\mathrm{OH}}^{-}\right]}^2$

$=\left[\mathrm{s}\right]{\left[2\mathrm{s}+0.1\right]}^2$

$=\left[\mathrm{s}\right]\left[4{\mathrm{s}}^2+0.01+0.4\mathrm{S}\right]$$=4{\mathrm{s}}^3+0.01\mathrm{s}+0.4{\mathrm{s}}^2$

Neglecting higher power of s,

$$\begin{gathered} 2.2\times {10}^{-20}=0.01\mathrm{s} \\ \mathrm{s}=2.2\times {10}^{-18}\mathrm{M} \end{gathered}$$

Therefore, Solubility of $\mathrm{Cu}{\left(\mathrm{OH}\right)}_2$ in $0.1\mathrm{M}$ aqueous $\mathrm{NaOH}$ solution is $2.2\times {10}^{-18}\mathrm{M}$