Solubility Product

Introduced to this in AP Chemistry

Useful for determining solubility of ionic solids in an aqueous solution

BaF${}_2$ (s) $\leftrightarrow$ Ba${}^{2+}$ (aq) + 2 F${}^{-}$ (aq)

K${}_{sp}$ = [F${}^{-}$]${}^2$[Ba${}^{2+}$]

Solubility vs. Solubility Product §

• K${}_{sp}$ = solubility product - has only one value for a specific temperature
• Solubility = equilibrium position concentration: [] = molarity
• K${}_{sp}$ and molar solubility are not the same thing

Solving Solubility Problems §

The salt, BaF${}_2$ at 25$°$ C has a measured solubility of 1.82 * 10${}^{-2}$ Calculate K${}_{sp}$

BaF${}_2$ (s) $\leftrightarrow$ Ba${}^{2+}$ + 2F${}^{-}$

Solving Solubility with a Common Ion §

For the salt Ag${}_2$CrO${}_4$ at 25$°$C, K${}_{sp}$ = 9.0 * 10${}^{-12}$

What is the solubility in 0.050 M Na${}_2$CrO${}_4$

Ag${}_2$CrO${}_4$ (s) $\leftrightarrow$	2Ag${}^{+}$ (aq)	CrO${}_4^{2-}$ (aq)
Initial	0	0.050
Change	+2s	+s
Equilibrium	2s	0.050+s

Predicting Formation of Precipitate §

Q${}_{sp}$ is ion product expressed in the same way as K${}_{sp}$ for a particular system

• K${}_{sp}$ = Q${}_{sp}$ $\to$ saturated solution, but no percipitation
• K${}_{sp}$ < Q${}_{sp}$ $\to$ saturated solution, with percipitation
• K${}_{sp}$ > Q${}_{sp}$ $\to$ unsaturated solution

Example §

If the solutions are added together, predict if a precipitate of PbCl${}_2$ will form. (PbCl${}_2$, K${}_{sp}$ = 1.6 * 10${}^{-5}$)

30.0 mL of 0.10 M NaCl

20.0 mL of 0.025 M Pb(NO${}_3$)${}_2$

[Pb${}^{2+}$] = $\frac{20.0mL\ast 0.025M}{50.0mL}$ = 0.010 M

[Cl${}^{-}$] = $\frac{30.0mL\ast 0.10M}{50.0mL}$ = 0.060 M

Q${}_{sp}$ = [Pb${}^{2+}$][Cl${}^{-}$]${}^2$ = (0.010 M)(0.060 M)${}^2$ = 3.6 * 10${}^{-5}$

K${}_{sp}$ = 1.6 * 10${}^{-5}$

K${}_{sp}$ < Q${}_{sp}$ $\to$ precipitate of PbCl${}_2$ will form