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 $\ce{Cu(OH)2}$?
No comments:
Post a Comment