Precipitation


We can predict when precipitation will occur by considering the equlibrium expression involving solids. For example, PbSO4 is slightly soluble in water.

\[\mathrm{PbSO_4}(s) \rightleftharpoons \mathrm{Pb^{2+}}(aq) + \mathrm{SO_4^{2-}}(aq) \qquad K_{\mathrm{sp}} = [\mathrm{Pb^{2+}}][\mathrm{SO_4^{2-}}] = 1.3\times 10^{-8} \]

Imagine having a 0.1 M aqueous sodium sulfate solution that was exactly 1.0 L.

\[\mathrm{Na_2SO_4}(s) \longrightarrow \mathrm{2Na^{+}}(aq) + \mathrm{SO_4^{2-}}(aq) \qquad K_{\mathrm{sp}} = [\mathrm{Na^{+}}]^2[\mathrm{SO_4^{2-}}] >> 1 \]

The solution would then contain a 0.1 M concentration of SO42– ions. Now let us introduce Pb2+ to the solution by adding highly soluble lead(II) acetate (Pb(CH3COOH)2; Ksp >> 1) compound that contains a common ion with PbSO4. (If you didn’t spot it, the common ion is Pb2+.)

How much Pb(CH3COOH)2 (in g) would we need to add to begin precipitating PbSO4?

We can determine what the Pb2+ concentration needs to be for precipitation of PbSO4 to start occuring. We will write an ICE table and then solve for the Pb2+ ion concentration in the equilibrium expression.

PbSO4(s) \(\rightleftharpoons\) Pb2+(aq) + SO42–(aq)
I ≈0 0.1
C +x +x
E x 0.1+x

\[\begin{align*} [\mathrm{Pb^{2+}}][\mathrm{SO_4^{2-}}] &= K_{\mathrm{sp}} \\[1ex] x(0.1+x) &=1.3\times 10^{-8}\\[1ex] 0.1x &= 1.3\times 10^{-8}\\[1ex] x &= 1.3\times 10^{-9} ~~ \overset{\mathrm{test}~x}{\longrightarrow} ~~ \dfrac{1.3\times 10^{-9}}{0.1}\times 100\% = 1.3\times 10^{-6}\% \\[1ex] [\mathrm{Pb}^{2+}]_{\mathrm{eq}} &= 1.3\times 10^{-9} \end{align*}\]

We now know that PbSO4 will begin to precipitate out of this solution if the Pb2+ concentration surpasses 1.3 × 10–9 M.

So, what amount (in g) of Pb(CH3COOH)2 would we need to add to our solution to reach our target Pb2+ concentration? Well, the minimum concentration of Pb(CH3COOH)2 would need to be 1.3 × 10–9 M. Convert this to mass. To make things simple, let us ignore the volume change of the solution by the addition of our solute (just assume the volume stays at 1.0 L).

\[\begin{align*} [\mathrm{Pb(CH_3COOH)_2}] \times V_{\mathrm{soln}} \times \mathrm{m.w.~of~Pb(CH_3COOH)_2} &= m_{\mathrm{Pb(CH_3COOH)_2}} \\[2ex] \left ( \dfrac{1.3\times 10^{-9}~\mathrm{mol~Pb(CH_3COOH)_2}}{1.0~\mathrm{L}} \right ) \left ( 1.0~\mathrm{L} \right ) \left ( \dfrac{325.29~\mathrm{g}}{\mathrm{mol}}\right ) &= 4.23\times 10^{-7}~\mathrm{g~Pb(CH_3COOH)_2} \end{align*}\]

We see that a very tiny amount of Pb(CH3COOH)2 is needed to begin precipitating PbSO4 from our solution.

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