# Why aren't the copper(II) ions in CuSO4 precipitated by hydroxide ions from water?

From my own experience and literature, I know that $\ce{CuSO4}$ is well soluble in water and dissociates into $\ce{Cu^2+}$ and $\ce{SO4^2-}$. $\ce{Cu(OH)2}$, however, is not: $K_\mathrm{sp} = 2 \cdot 10^{-19}$.

Therefore, in pure water which has a pH of $7$, $c(\ce{OH-}) = 10^{-7}$. This means that a concentration of $2 \cdot 10^{-19} / (10^{-7})^2 = 2 \cdot 10^{-5}$ of $\ce{Cu^2+}$ would be enough to cause precipitation. How come, when dissolving $\ce{CuSO4}$, the $\ce{Cu^2+}$ stays in solution, and doesn't precipitate with the $\ce{OH-}$ ions of water to form $\ce{Cu(OH)2}$?

• Copper (II) sulfate forms a copper aquo complex $\ce{Cu(H2O)6^2+}$ which is slightly acidic. See: en.wikipedia.org/wiki/Metal_aquo_complex – MaxW Dec 5 '15 at 17:05
• Long story short, $\ce{Cu^2+}$ is slightly hydrolyzed and hence acidic. If you'd try to make it exactly neutral, a precipitate would form. – Ivan Neretin Dec 5 '15 at 23:40

In aqueous solution, the copper(II) ion forms the aqua complex $\ce{[Cu(H2O)6]^2+}$:

$\hspace{55 mm}$

Because the central $\ce{Cu^2+}$ ion is positively charged, it polarises the $\ce{O-H}$ bonds towards oxygen, which makes the hydrogens more acidic. Therefore, the ion acts as a weak Brønsted acid, with a $\mathrm{p}K_\mathrm{a}$ of approximately $8$ (J. Phys. Chem. A, 2015, 119 (12), pp 2926–2939):

$$\ce{[Cu(H2O)6]^2+ <=> H+ + [Cu(H2O)5(OH)]^+}$$

Suppose we have a $0.1~\mathrm{mol~dm^{-3}}$ solution of $\ce{CuSO4}$. To a first approximation,

\begin{align} [\ce{Cu(H2O)6^2+}] &= 0.1~\mathrm{M} \\ [\ce{H+}] &= \sqrt{(0.1~\mathrm{M})(10^{-8}~\mathrm{M})} \\ &= 3.162 \times 10^{-5}~\mathrm{M} \\ [\ce{OH-}] &= \frac{10^{-14}~\mathrm{M^2}}{[\ce{H+}]} \\ &= 3.162 \times 10^{-10}~\mathrm{M} \end{align}

Therefore,

\begin{align} [\ce{Cu^2+}][\ce{OH-}]^2 &= (0.1~\mathrm{M})(3.162\times 10^{-10}~\mathrm{M})^2 \\ &= 10^{-20}~\mathrm{M^3} < K_{\mathrm{sp}} \end{align}

and no precipitation of $\ce{Cu(OH)2}$ occurs.

If the pH of a $\ce{Cu^2+}$ solution is neutralised by careful addition of base, then precipitation will be observed as Ivan noted in his comment. The exact pH at which precipitation will occur is of course dependent on the concentration of $\ce{Cu^2+}$. The same argument here actually applies to all of the transition metals: all of them form rather insoluble hydroxides, but they also form acidic aqua complexes, which prevent the precipitation of the hydroxide.

The calculations could have been done more accurately by introducing fewer assumptions; however, they should not affect the final conclusion. (If this was a textbook, it would probably say: this exercise is left to the reader!)