Chapter U6.6, Problem 6E

Interpretation Introduction

Interpretation:

The method to improve the solubility of solid Cu(OH)₂ in an aqueous solution needs to be explained.

Concept introduction:

The strong acid, proton $\left(H^{+}\right)$ neutralizes the strong base, hydroxide $\left({\text{OH}}^{\text{-}}\right)$ in an acid-base neutralization reaction as per the equation below. Water, H₂O is produced by the reaction which remains unionized in an aqueous solution.

${\text{H}}^{+}\left(aq\right)+{\text{ OH}}^{-}\left(aq\right)\to {\text{H}}_{\text{2}}\text{O }\left(l\right)$

The solubility of copper hydroxide, Cu(OH)₂ in an aqueous solution may be improved by using the above acid-base reaction.

Answer to Problem 6E

The solubility of Cu(OH)₂ in water can be improved by adding an acid, viz., hydrochloric acid, HCl to the solution.

Explanation of Solution

The solubility of Cu(OH)₂ in water is given as per the reaction below:

$\text{Cu}{\left(\text{OH}\right)}_{\text{2}}\left(s\right)\rightleftharpoons {\text{ Cu}}^{2+}\left(aq\right)+2{\text{ OH}}^{-}\left(aq\right)$

The solubility product constant, K_sp for the solubility equilibrium above is given as

$K_{sp}=\left[{\text{Cu}}^{2+}\right]{\left[{\text{OH}}^{-}\right]}^2$

Where the square braces, [..] denote the molar concentrations of the aqueous species at equilibrium.

K_sp is a thermodynamic equilibrium constant and hence, must remain constant at a particular temperature.

Hydrochloric acid, HCl is a strong electrolyte in aqueous solution and ionizes as per the equation below.

$\text{HCl }\left(aq\right)\to {\text{H}}^{+}\left(aq\right)+{\text{ Cl}}^{-}\left(aq\right)$

The addition of HCl to the aqueous solution produces protons, ${\text{H}}^{+}$ combines with the ${\text{OH}}^{-}$ to give H₂O which remains undissociated in water.

${\text{H}}^{+}\left(aq\right)+{\text{ OH}}^{-}\left(aq\right)\to {\text{H}}_{\text{2}}\text{O }\left(l\right)$

The above reaction removes ${\text{H}}^{+}$ from the solution as H₂O. The removal of ${\text{H}}^{+}$ tends to decrease $\left[{\text{H}}^{+}\right]$ and hence, K_sp. However, K_sp being a thermodynamic equilibrium constant, must remain unaltered at the given temperature. To keep K_sp constant, $\left[{\text{Cu}}^{2+}\right]$ must increase and this is possible only when more Cu(OH)₂ dissolves in water, i.e., the addition of HCl increases the solubility of Cu(OH)₂ in water.

Conclusion

The addition of a strong acid will increase the solubility of Cu(OH)₂ in water.