Question

The pH of a 0.2 M solution of CuSO₄ is 4.0. Write chemical equations to explain why a solution of this salt is neither basic [from the reaction of SO₄^2- with water] nor neutral but acidic.

Short answer

$S{O_4}^{2-}0\left(aq\right)+2H^{+}\left(aq\right)\to H_{2}S{O}_4\left(aq\right)$

Step-by-step solution

The pH of a solution

A solution's pH represents its acidic, basic, or neutral nature.

$$\begin{gathered} \mathit{I}\mathit{f}\mathbf{,}\mathbf{}\mathit{p}\mathit{H}\mathbf{<}\mathbf{7}\mathbf{,}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathit{n}\mathbf{}\mathit{a}\mathit{c}\mathit{i}\mathit{d}\mathit{i}\mathit{c}\mathbf{}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n} \\ \mathit{p}\mathit{H}\mathbf{=}\mathbf{7}\mathbf{,}\mathbf{}\mathit{n}\mathit{e}\mathit{u}\mathit{t}\mathit{r}\mathit{a}\mathit{l}\mathbf{}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{u}\mathit{i}\mathit{o}\mathit{n} \\ \mathit{p}\mathit{H}\mathbf{>}\mathbf{7}\mathbf{,}\mathbf{}\mathit{b}\mathit{a}\mathit{s}\mathit{i}\mathit{c}\mathbf{}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{.} \end{gathered}$$

Explanation

The salt CuSO₄ is formed by combining a weak base Cu(OH)₂ and a strong acid H₂SO₄.

$H_{2}S{O}_4\left(aq\right)+Cu{\left(OH\right)}_2\to CuS{O}_4\left(aq\right)+2H_{2}O\left(I\right)$

Thus, when CuSO₄ dissolved in water, the resultant solution is acidic due to the formation of H₂SO₄ a strong acid.

$S{O_4}^{2-}\left(aq\right)+2H^{+}\left(aq\right)\to H_{2}S{O}_4\left(aq\right)$