Question 17.SE.16: Does a precipitate form when 0.10 L of 8.0 × 10^-3 M Pb(NO3)...

Analyze The problem asks us to determine whether a precipitate forms when two salt solutions are combined.

Plan We should determine the concentrations of all ions just after the solutions are mixed and compare the value of Q with K$_{sp}$ for any potentially insoluble product. The possible metathesis products are PbSO$_4$ and NaNO$_3$. Like all sodium salts $NaNO_3$ is soluble, but PbSO$_4$ has a K$_{sp}$ of 6.3 × 10$^{-7}$ (Appendix D) and will precipitate if the Pb$^{2+}$ and $SO_4^{2-}$ concentrations are high enough for Q to exceed K$_{sp}$.

Solve
When the two solutions are mixed, the volume is 0.10 L + 0.40 L = 0.50 L. The number of moles of Pb$^{2 +}$ in 0.10 L of 8.0 × 10$^{-3}$M $Pb(NO_3)_2$ is:

$(0.10\, \cancel{L} )\left(\frac{8.0 \times 10^{-3}\,mol}{\cancel{L}}\right)=8.0 \times 10^{-4}\,mol$

The concentration of Pb$^{2 +}$ in the 0.50 L mixture is therefore:

$\left[ Pb ^{2+}\right]=\frac{8.0 \times 10^{-4}\,mol}{0.50 \,L}=1.6 \times 10^{-3}\,M$

The number of moles of $SO _4^{2-}$ in 0.40 L of 5.0 × 10$^{-3}$ M $Na_2SO_4$ is:

$(0.40\, \cancel{L} )\left(\frac{5.0 \times 10^{-3}\,mol}{\cancel{L}}\right)=2.0 \times 10^{-3}\,mol$

Therefore:

$\left[ SO _4{}^{2-}\right]=\frac{2.0 \times 10^{-3}\,mol}{0.50 \,L}=4.0 \times 10^{-3}\,M$

and:

$Q=\left[ Pb ^{2+}\right]\left[ SO _4^{2-}\right]=\left(1.6 \times 10^{-3}\right)\left(4.0 \times 10^{-3}\right)=6.4 \times 10^{-6}$

Because Q >K$_{sp}$, $PbSO_4$ precipitates.