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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.

CowperKettle
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    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 03 '16 at 17:59
  • @IvanNeretin $\ce{HSO4-}$ isn't a strong acid; $\mathrm pK_\mathrm a \approx 2$. – hBy2Py May 03 '16 at 18:15
  • Still, that's strong enough to make the overall result acidic. – Ivan Neretin May 03 '16 at 18:17
  • @IvanNeretin Do you know the magnitude of the $\ce{Cu^{2+}}\mathrm{-}\ce{OH-}$ stability constants offhand? In general I agree, it's likely to be acidic, but I wonder if it's a closer thing than it first seems. – hBy2Py May 03 '16 at 18:21
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    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 03 '16 at 18:29
  • @Jan Wouldn't $\ce{Cu(OH)2}$ precipitation and higher hydrolysis forms potentially also need to be considered? (They might prove negligible, yes, but...) – hBy2Py May 03 '16 at 18:31
  • @Brian I’m pretty sure that they can be neglected at first approximation (and possibly even at second) because of the significantly higher $\mathrm{p}K_\mathrm{a}$ value of $\ce{[Cu(OH)(H2O)5]+}$. But as I learnt from aluminium complexes there might actually be some fascinating chemistry in there. – Jan May 03 '16 at 18:33
  • @Jan Exactly why I'm pressing the issue. It could be just as boring as you and Ivan have laid out. But, it might not be. I have some literature on which to fiddle -- I will probably be posting an answer either way sometime soon. – hBy2Py May 03 '16 at 18:38
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    @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 04 '16 at 17:25

1 Answers1

7

Because the second pKa of sulfuric acid is weak, sulfate withdraws protons from water. Therefore, the contribution of sulfate alone is to make the solution slightly more basic than pH 7.

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.

DavePhD
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