Solubility and pH for metal hydroxides
For sparingly soluble hydroxides, solubility depends strongly on pH. The tool models:
\[
\mathrm{M(OH)_n}(s) \rightleftharpoons
\mathrm{M^{n+}}(aq) + n\,\mathrm{OH^-}(aq)
\]
with
\[
K_{sp} = [\mathrm{M^{n+}}]\,[\mathrm{OH^-}]^{\,n}
\]
At \(25^\circ\text{C}\): \(K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0\times10^{-14}\), so
\[
\mathrm{pH} = -\log_{10}[\mathrm{H_3O^+}],\quad
\mathrm{pOH} = -\log_{10}[\mathrm{OH^-}],\quad
\mathrm{pH} + \mathrm{pOH} = 14.00.
\]
Ion product \(Q_{sp}\) and precipitation
For a solution with known \([\mathrm{M^{n+}}]_0\) and pH (or pOH), the ion product is
\[
Q_{sp} = [\mathrm{M^{n+}}]_0\,[\mathrm{OH^-}]^{\,n}.
\]
- \(Q_{sp} < K_{sp}\): unsaturated, no precipitation.
- \(Q_{sp} \approx K_{sp}\): saturated, at equilibrium.
- \(Q_{sp} > K_{sp}\): supersaturated, precipitation occurs.
In “Check precipitation at given pH” mode the calculator:
- Converts pH or pOH to \([\mathrm{OH^-}]\) using \(K_w\).
- Computes \(Q_{sp} = [\mathrm{M^{n+}}]_0[\mathrm{OH^-}]^{n}\).
- Compares \(Q_{sp}\) with \(K_{sp}\) to decide if precipitation occurs.
-
If \(Q_{sp} > K_{sp}\), uses
\[
[\mathrm{M^{n+}}]_{\text{eq}} = \frac{K_{sp}}{[\mathrm{OH^-}]^{n}}
\]
to estimate the remaining dissolved metal and the fraction precipitated.
Threshold pH (onset of precipitation)
At the pH where precipitation just begins, only a tiny amount has precipitated, so
\([\mathrm{M^{n+}}]_{\text{crit}} \approx [\mathrm{M^{n+}}]_0\). Then
\[
K_{sp} \approx [\mathrm{M^{n+}}]_0\,[\mathrm{OH^-}]_{\text{crit}}^{\,n}
\quad\Rightarrow\quad
[\mathrm{OH^-}]_{\text{crit}} =
\left(\frac{K_{sp}}{[\mathrm{M^{n+}}]_0}\right)^{1/n}.
\]
From this the tool finds \(\mathrm{pOH}_{\text{crit}}\) and
\(\mathrm{pH}_{\text{crit}} = 14 - \mathrm{pOH}_{\text{crit}}\).
For pH values higher than \(\mathrm{pH}_{\text{crit}}\), the hydroxide becomes less soluble
and precipitation is favored.
pH is treated as known (e.g. from a buffer or separate acid–base calculation). All relations
assume \(25^\circ\text{C}\) so that \(K_w = 1.0\times10^{-14}\).