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}$$ at room temperature.)

Calculate the final concentrations and compare Q$$_{sp}$$ 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}$$ at room temperature.)

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

Accepted Answer

Yes, a precipitate forms because $$Q_{sp} \u003e K_{sp}$$ after mixing.

Predicting Precipitation Using Ksp and Solution Mixing

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