6.1 Solubility and Equilibrium
We can apply standard equilibrium problem techniques to solving equilibrium concentrations of soluble compounds in water.
6.1.1 Example 1: AgCl
Determine the molar concentration and the mass concentration (in g L–1) of silver(I) chloride (m.m. = 143.32 g mol–1) in a saturated aqueous solution.
\[\mathrm{AgCl}(s) \rightleftharpoons \mathrm{Ag^+}(aq) + \mathrm{Cl^-}(aq) \quad K_{\mathrm{sp}} = 1.77\times 10^{-10}\]
If the volume of the solution was 5.0 L, determine the mass (in g) of AgCl in this solution.
Set up the ICE table.
| AgCl(s) | ⇌ | Ag+(aq) | + | Cl–(aq) | |
|---|---|---|---|---|---|
I |
0 |
0 | |||
C |
+x |
+x | |||
E |
x |
x |
Write out the equilibrium expression and then solve for x.
\[\begin{align*} [\mathrm{Ag^+}][\mathrm{Cl^-}] &= K_{\mathrm{sp}}\\ (x)(x) &= 1.77\times 10^{-10}\\ x^2 &= 1.77\times 10^{-10}\\ x &= 1.33\times 10^{-5}\\[2ex] [\mathrm{Ag^+}]_{\mathrm{eq}} &= 1.33\times 10^{-5}~M \\ [\mathrm{Cl^-}]_{\mathrm{eq}} &= 1.33\times 10^{-5}~M \end{align*}\]
Determine the molar concentration of the reactant by using stoichiometry and one of the products. Here I choose to use Ag+ for this conversion.
\[\begin{align*} [\mathrm{AgCl}] &= \dfrac{1.33\times 10^{-5}~\mathrm{mol~Ag^+}}{\mathrm{L}} \left ( \dfrac{1~\mathrm{mol~AgCl}}{1~\mathrm{mol~Ag^+}} \right ) \\[1.5ex] &= 1.33\times 10^{-5}~M \end{align*}\]
Therefore, this saturated solution is a 1.33 × 10–5 M AgCl aqueous solution.
To determine the mass concentration (in g L–1), multiply the molar concentration by the molar mass of the compound.
\[\begin{align*} \rho_{\mathrm{AgCl}} &= \left ( 1.33\times 10^{-5}~\mathrm{mol~L^{-1}} \right ) \left ( 143.32~\mathrm{g~mol^{-1}}\right ) \\[1.5ex] &= 1.91\times 10^{-3}~\mathrm{g~L^{-1}} \end{align*}\]
Therefore, this saturated solution is a 1.91 × 10–3 g L–1 AgCl aqueous solution. The solution is very dilute (not much AgCl dissolves in water; it is very slightly soluble).
Finally, determine the mass of AgCl (in g) in a 5.0 L saturated solution. Simply multiply the mass concentration by the volume of solution.
\[\begin{align*} m_{\mathrm{AgCl}} &= \left (1.91\times 10^{-3}~\mathrm{g~L^{-1}}\right ) \left ( 5.0~\mathrm{L} \right ) \\ &= 9.55\times 10^{-3}~\mathrm{g} \end{align*}\]
6.1.2 Exmaple 2: PbCl2
Determine the molar concentration and the mass concentration (in g L–1) of lead(II) chloride (m.m. = 278.1 g mol–1) in a saturated aqueous solution.
\[\mathrm{PbCl_2}(s) \rightleftharpoons \mathrm{Pb^{2+}}(aq) + \mathrm{2Cl^-}(aq) \quad K_{\mathrm{sp}} = 1.7\times 10^{-5}\]
If the volume of the solution was 2.0 L, determine the mass (in g) of PbCl2 in this solution.
Set up the ICE table.
| PbCl2(s) | ⇌ | Pb2+(aq) | + | 2Cl–(aq) | |
|---|---|---|---|---|---|
I |
0 |
0 | |||
C |
+x |
+2x | |||
E |
x |
2x |
Write out the equilibrium expression and then solve for x.
\[\begin{align*} [\mathrm{Pb^{2+}}][\mathrm{Cl^-}]^2 &= K_{\mathrm{sp}}\\ (x)(2x)^2 &= 1.7\times 10^{-5}\\ 4x^3 &= 1.7\times 10^{-5}\\ x &= 1.62\times 10^{-2}\\[2ex] [\mathrm{Pb^{2+}}]_{\mathrm{eq}} &= 1.62\times 10^{-2}~M \\ [\mathrm{Cl^-}]_{\mathrm{eq}} &= 2(1.62\times 10^{-2}~M) = 3.23\times 10^{-2}~M \end{align*}\]
Determine the molar concentration of the reactant by using stoichiometry and one of the products. Here I choose to use Pb2+ for this conversion.
\[\begin{align*} [\mathrm{PbCl_2}] &= \dfrac{1.62\times 10^{-2}~\mathrm{mol~Pb^{2+}}}{\mathrm{L}} \left ( \dfrac{1~\mathrm{mol~PbCl_2}}{1~\mathrm{mol~Pb^{2+}}} \right ) \\[1.5ex] &= 1.62\times 10^{-2}~M \end{align*}\]
Therefore, this saturated solution is a 1.62 × 10–2 M PbCl2 aqueous solution.
To determine the mass concentration (in g L–1), multiply the molar concentration by the molar mass of the compound.
\[\begin{align*} \rho_{\mathrm{PbCl_2}} &= \left ( 1.62\times 10^{-2}~\mathrm{mol~L^{-1}} \right ) \left ( 278.1~\mathrm{g~mol^{-1}}\right ) \\[1.5ex] &= 4.50~\mathrm{g~L^{-1}} \end{align*}\]
Therefore, this saturated solution is a 4.50 g L–1 PbCl2 aqueous solution.
Finally, determine the mass (in g) of PbCl2 in a 2.0 L saturated solution. Simply multiply the mass concentration by the volume of solution.
\[\begin{align*} m_{\mathrm{PbCl_2}} &= \left ( 4.50~\mathrm{g~L^{-1}} \right ) \left ( 2.0~\mathrm{L} \right ) \\ &= 9.0~\mathrm{g} \end{align*}\]
6.1.3 Exmaple 3: Ag2SO4
Determine the molar concentration and the mass concentration (in g L–1) of silver sulfate (Ag2SO4; m.m. = 311.799 g mol–1) in a saturated aqueous solution.
\[\mathrm{Ag_2SO_4}(s) \rightleftharpoons \mathrm{2Ag^{+}}(aq) + \mathrm{SO_4^{2-}}(aq) \quad K_{\mathrm{sp}} = 1.2\times 10^{-5}\]
If the volume of the solution was 750.0 mL, determine the mass (in g) of Ag2SO4 in this solution.
Set up the ICE table.
| Ag2SO4(s) | ⇌ | 2Ag+(aq) | + | SO42–(aq) | |
|---|---|---|---|---|---|
I |
0 |
0 | |||
C |
+2x |
+x | |||
E |
2x |
x |
Write out the equilibrium expression and then solve for x.
\[\begin{align*} [\mathrm{Ag^{+}}]^2[\mathrm{SO_4^{2-}}] &= K_{\mathrm{sp}}\\ (2x)^2(x) &= 1.2\times 10^{-5}\\ 4x^3 &= 1.2\times 10^{-5}\\ x &= 1.44\times 10^{-2}\\[2ex] [\mathrm{Ag^{+}}]_{\mathrm{eq}} &= 2(1.44\times 10^{-2}~M) = 2.88\times 10^{-2}~M\\ [\mathrm{SO_4^{2-}}]_{\mathrm{eq}} &= 1.44\times 10^{-2}~M \end{align*}\]
Determine the molar concentration of the reactant by using stoichiometry and one of the products. Here I choose to use SO42– for this conversion.
\[\begin{align*} [\mathrm{Ag_2SO_4}] &= \dfrac{1.44\times 10^{-2}~\mathrm{mol~SO_4^{2+}}}{\mathrm{L}} \left ( \dfrac{1~\mathrm{mol~Ag_2SO_4}}{1~\mathrm{mol~SO_4^{2-}}} \right ) \\[1.5ex] &= 1.44\times 10^{-2}~M \end{align*}\]
Therefore, this saturated solution is a 1.44 × 10–2 M Ag2SO4 aqueous solution.
To determine the mass concentration (in g L–1), multiply the molar concentration by the molar mass of the compound.
\[\begin{align*} \rho_{\mathrm{Ag_2SO_4}} &= \left ( 1.44\times 10^{-2}~\mathrm{mol~L^{-1}} \right ) \left ( 311.799~\mathrm{g~mol^{-1}}\right ) \\[1.5ex] &= 4.50~\mathrm{g~L^{-1}} \end{align*}\]
Therefore, this saturated solution is a 4.50 g L–1 Ag2SO4 aqueous solution.
Finally, determine the mass (in g) of Ag2SO4 in a 750.0 mL saturated solution. Simply multiply the mass concentration by the volume of solution.
\[\begin{align*} m_{\mathrm{Ag_2SO_4}} &= \left ( 4.50~\mathrm{g~L^{-1}} \right ) \left ( 0.750~\mathrm{L} \right ) \\ &= 3.375~\mathrm{g} \end{align*}\]