General Chemistry 2: Solubility Product

By experiment, it is found that of 6.2 x 10$^{-12}$ moles of Pb$_3$(PO$_4$)$_2$ dissolves in 1 L of aqueous solution at 25$^\circ$C. What is the solubility product constant, K$_{sp}$, at this temperature?

Calculate the solubility (in grams per liter) of PbI$_2$ in water from the solubility product constant 1.4 x 10$^{-8}$.

What are the concentrations of the ionic species in a saturated solution of Ag$_2$CrO$_4$

K$_{sp}$ = 1.1 x 10$^{-12}$

Calculate the equilibrium concentration of Pb$^{2+}$, I$^-$, Na(\^+$, and NO$_3^-$when 100.0 mL of 0.0500 M Pb(NO$_3$)$_2## and 200.0 mL of 0.100 M NaI are mixed.

PbI$_2$ K$_{sp}$ = 1.4 x 10$^{-8}$

What is the molar solubility of Ag$_2$CrO$_4$ in 0.100 M AgNO$_3$? Compare this to the molar solubility of Ag$_2$CrO$_4$ in water, 1.3 x 10$^{-4}$ M

Ag$_2$CrO$_4$ K$_{sp}$ = 9.0 x 10$^{-12}$

The concentration of Mg(NO$_3$)$_2$ is 4 x 10$^{-4}$ M and the concentration of NaOH is 2 x 10$^{-4}$, do you expect Mg(OH)$_2$ to precipitate?

Mg(OH)$_2$ K$_{sp}$ = 8.9 x 10$^{-12}$

A solution contains 1.0 x 10$^{-4}$ M Cu$^+$ and 2.0 x 10$^{-3}$ M Pb$^{2+}$. When I$^{-}$ is added, which will precipitate first? How much I$^-$ is needed to cause a precipitate to form?When CuI is added to PbI$_2$

CuI K$_{sp}$ = 5.3 x 10$^{-12}$

PbI$_2$ K$_{sp}$ = 1.4 x 10$^{-8}$

What is the solubility of AgCl in 0.020 M NaCN?

AgCl K$_{sp}$ = 1.8 x 10$^{-10}$

Ag(CN)$_2^-$ K$_{f}$ = 5.6 x 10$^{18}$

0.25 moles of AgNO$_3$ are dissolved in 1 L of 0.60 M NaCN and 1 mole of NaCl is added. Will AgCl precipitate?

AgCl K$_{sp}$ = 1.8 x 10$^{-10}$

Ag(CN)$_2^-$ K$_{f}$ = 5.6 x 10$^{18}$

How many moles of Mg(OH)$_2$ can be dissolved in a 1 L solution that already contains 0.52 moles Mg$^{2+}$.

Mg(OH)$_2$ K$_{sp}$ = 8.9 x 10$^{-12}$

The solubility of Ce(IO$_3$)$_3$ in a 0.20 M KIO$_3$ solution is 4.4 x 10$^{-8}$ mol/L. Calculate K$_{sp}$ for Ce(IO$_3$)$_3$.

Calculate the concentrations of Ag$^+$, Ag(S$_2$O$_3$)$^-$, and Ag(S$_2$O$_3$)$_2^{3-}$ in a solution prepared by mixing 150.0 mL of 1.00 x 10$^{-3}$ M AgNO$_3$ with 200.0 mL of 5.00 M Na$_2$S$_2$O$_3$. The stepwise formation equilibria are:

Ag$^+$ + S$_2$O$_3^{2-}$ $\leftrightharpoons$ Ag(S$_2$O$_3$)$^-$ K$_{f1}$ = 7.4 x 10$^{8}$

Ag(S$_2$O$_3$)$^-$ + S$_2$O$_3^{2-}$ $\leftrightharpoons$ Ag(S$_2$O$_3$)$_2^{3-}$ K$_{f2}$ = 3.9 x 10$^{4}$