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.