# Why will CuSO4 solution be acidic?

An end-chapter problem posed in an 11-grade texbook asks to analyse whether a water solution of CuSO4 will be acidic.

The standard answer is yes, because "$\ce{H2SO4}$ is a strong acid, hence $\ce{SO_4^2-}$ is a weak base. As a weak base, it will not be too hot about grabbing $\ce{H^+}$ from the environment. On its part, $\ce{Cu^2+}$ will strongly attract $\ce{OH^-}$ groups, and overall protons will prevail over hydroxides."

But I was told in the chat that the real answer is more interesting and complicated, so I post this question in the hope that someone will explain the process in-depth.

• There is nothing complicated about it. A salt formed by a strong acid and a weak base will make the solution acidic due to hydrolysis. – Ivan Neretin May 3 '16 at 17:59
• @IvanNeretin $\ce{HSO4-}$ isn't a strong acid; $\mathrm pK_\mathrm a \approx 2$. – hBy2Py May 3 '16 at 18:15
• Still, that's strong enough to make the overall result acidic. – Ivan Neretin May 3 '16 at 18:17
• The question does indeed boil down to the $\mathrm{p}K_\mathrm{a}$ value of $\ce{[Cu(H2O)6]^2+}$, which I don’t know off the top of my head but should be greater than seven. The remainder is basic maths. – Jan May 3 '16 at 18:29
• @Jan You're right, it's pretty boring. The various hydroxyl complexes only come into play if you add base. DavePhD's answer is pretty much all there is to it. – hBy2Py May 4 '16 at 17:25

However, as quantified in CE4501 Environmental Engineering Chemical Processes, Cu will form species such as CuOH+, which is the reason the solution will be acidic overall. This could be approximated by thinking of $\ce{Cu(H2O)6^{2+}}$ as having a pKa of ~6.3.
Because $\ce{HSO4-}$ is a stronger acid than CuOH+ is a base, the solution is acidic.