9
$\begingroup$

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.

Improve this question
$\endgroup$
9
  • 2
    $\begingroup$ 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. $\endgroup$
    – Ivan Neretin
    Commented May 3, 2016 at 17:59
  • $\begingroup$ @IvanNeretin $\ce{HSO4-}$ isn't a strong acid; $\mathrm pK_\mathrm a \approx 2$. $\endgroup$
    – hBy2Py
    Commented May 3, 2016 at 18:15
  • $\begingroup$ Still, that's strong enough to make the overall result acidic. $\endgroup$
    – Ivan Neretin
    Commented May 3, 2016 at 18:17
  • 3
    $\begingroup$ 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. $\endgroup$
    – Jan
    Commented May 3, 2016 at 18:29
  • 1
    $\begingroup$ @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. $\endgroup$
    – hBy2Py
    Commented May 4, 2016 at 17:25

1 Answer 1

Reset to default
7
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.