Does a precipitate form when $$0.10\ \mathrm{L} of$$ $$8.0 imes 10$$^{-3}$$\ \mathrm{M}\ \mathrm{Pb(NO_3$)_2} is $$added to $$0.40\ \mathrm{L} of$$ $$5.0 imes 10$$^{-3}$$\ \mathrm{M}\ \mathrm{Na_2SO_4}$$? (Assume $$K_{sp}$$ for $$\mathrm{$$PbSO_4$$} is$$ $$1.6 imes 10$$^{-8}$$.)

Calculate the final concentrations, the ion product Q$, and compare to K$$_{sp}$$ to justify your answer.

Question

Does a precipitate form when $$0.10\ \mathrm{L} of$$ $$8.0 	imes 10$$^{-3}$$\ \mathrm{M}\ \mathrm{Pb(NO_3$)_2} is $$added to $$0.40\ \mathrm{L} of$$ $$5.0 	imes 10$$^{-3}$$\ \mathrm{M}\ \mathrm{Na_2SO_4}$$? (Assume $$K_{sp}$$ for $$\mathrm{$$PbSO_4$$} is$$ $$1.6 	imes 10$$^{-8}$$.)

Calculate the final concentrations, the ion product Q$, and compare to K$$_{sp}$$ to justify your answer.

Accepted Answer

Yes, a precipitate forms because $$Q \u003e K_{sp}$$.

$$\mathrm{[Pb$$^{2+}$$]_{final}} = \frac{(0.10\ \mathrm{L}) 	imes (8.0 	imes 10$$^{-3}$$\ \mathrm{M})}{0.50\ \mathrm{L}} = 1.6 	imes 10$$^{-3}$$\ \mathrm{M}$$

$$\mathrm{[SO_4$^{2-}]_{final}} = \frac{(0.40\ \mathrm{L}) 	imes (5.0 	imes 10$$^{-3}$$\ \mathrm{M})}{0.50\ \mathrm{L}} = 4.0 	imes 10$$^{-3}$$\ \mathrm{M}$$

$$Q = [\mathrm{Pb$$^{2+}$$}][\mathrm{SO_4$^{2-}}] = (1.6 	imes 10$$^{-3}$$)(4.0 	imes 10$$^{-3}$$) = 6.4 	imes 10$$^{-6}$$

Since Q$$ \u003e K_{sp}$$, a precipitate forms.

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