Solubility Product
Dissolving a solid is more complicated than it seems. At first glance, the dissolution of sodium chloride in water seems very straightforward. The salt dissociates into its constituent cations and anions, assuming complete dissociation.
\[\mathrm{NaCl}(s) \overset{\mathrm{H_2O}}\rightleftharpoons \mathrm{Na^+}(aq) + \mathrm{Cl^-}(aq)\]
Now, consider lead(II) chloride (PbCl2), a slightly soluble salt. When dissolved in water, three different equilibria occur simultaneously, each with their own equilibrium constant.
The solubility product for a substance is the equilibrium constant of the solid undergoing complete dissociation (i.e. dissolution) in water.
\[\mathrm{PbCl_2}(s) \overset{\mathrm{H_2O}}\rightleftharpoons \mathrm{Pb^{2+}}(aq) + \mathrm{2Cl^-}(aq) \quad K_{\mathrm{sp}} = 1.7\times 10^{-5}\]
Ksp values can exist for other solvents.
The solubility product, Ksp, for a substance is related to its solubility. If Ksp is > 1, the substance is quite soluble. If Ksp < 1, the substance is not very soluble.
Solubility and Equilibrium
We can apply standard equilibrium problem techniques to solving equilibrium concentrations of soluble compounds in water.
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.
\[\begin{align*} m_{\mathrm{AgCl}} &= \left (1.91\times 10^{-3}~\mathrm{g~L^{-1}}\right ) \left ( 5.0~\mathrm{L} \right ) \\ &= 9.55\times 10^{-3}~\mathrm{g} \end{align*}\]
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.
\[\begin{align*} m_{\mathrm{PbCl_2}} &= \left ( 4.50~\mathrm{g~L^{-1}} \right ) \left ( 2.0~\mathrm{L} \right ) \\ &= 9.0~\mathrm{g} \end{align*}\]
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.
\[\begin{align*} m_{\mathrm{Ag_2SO_4}} &= \left ( 4.50~\mathrm{g~L^{-1}} \right ) \left ( 0.750~\mathrm{L} \right ) \\ &= 3.375~\mathrm{g} \end{align*}\]
Common Ion
Solutions already containing an ion that is in common with the substance being dissolved naturally lowers the solubility of the starting substance.
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 0.1 M NaCl 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. Given that NaCl is dissolved in solution, the initial concentration of Cl–, the common ion, is 0.1 M.
| AgCl(s) | ⇌ | Ag+(aq) | + | Cl–(aq) | |
|---|---|---|---|---|---|
I |
0 |
0.1 | |||
C |
+x |
+x | |||
E |
x |
0.1+x |
Write out the equilibrium expression and then solve for x.
\[\begin{align*} [\mathrm{Ag^+}][\mathrm{Cl^-}] &= K_{\mathrm{sp}}\\ (x)(0.1 + x) &= 1.77\times 10^{-10}\\ 0.1x &= 1.77\times 10^{-10}\\ x &= 1.77\times 10^{-9} ~(<< 5\%)\\[2ex] [\mathrm{Ag^+}]_{\mathrm{eq}} &= 1.77\times 10^{-9}~M \\ [\mathrm{Cl^-}]_{\mathrm{eq}} &\approx 0.1~M \end{align*}\]
Determine the molar concentration of the reactant by using stoichiometry and a product that is not a common ion (here Ag+).
\[\begin{align*} [\mathrm{AgCl}] &= \dfrac{1.77\times 10^{-9}~\mathrm{mol~Ag^+}}{\mathrm{L}} \left ( \dfrac{1~\mathrm{mol~AgCl}}{1~\mathrm{mol~Ag^+}} \right ) \\[1.5ex] &= 1.77\times 10^{-9}~M \end{align*}\]
Therefore, this saturated solution is a 1.77 × 10–9 M AgCl aqueous solution. The solubility of AgCl is much lower in a NaCl solution than pure water!
To determine the mass concentration (in g L–1), multiply the molar concentration by the molar mass of the compound.
\[\begin{align*} \rho_{\mathrm{AgCl}} &= \left ( 1.77\times 10^{-9}~\mathrm{mol~L^{-1}} \right ) \left ( 143.32~\mathrm{g~mol^{-1}}\right ) \\[1.5ex] &= 2.54\times 10^{-7}~\mathrm{g~L^{-1}} \end{align*}\]
Therefore, this saturated solution is a 2.54 × 10–7 g L–1 AgCl aqueous solution.
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.
\[\begin{align*} m_{\mathrm{AgCl}} &= \left (2.54\times 10^{-7}~\mathrm{g~L^{-1}} \right ) \left ( 5.0~\mathrm{L} \right ) \\ &= 1.27\times 10^{-6}~\mathrm{g} \end{align*}\]
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 0.1 M NaCl 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 |
0.1 | |||
C |
+x |
+2x | |||
E |
x |
0.1+2x |
Write out the equilibrium expression and then solve for x.
Note: In this example, we end up with a cubic function. See the calculation performed using WolframAlpha.
\[\begin{align*} [\mathrm{Pb^{2+}}][\mathrm{Cl^-}]^2 &= K_{\mathrm{sp}}\\ (x)(0.1 + 2x)^2 &= 1.7\times 10^{-5}\\ (x)(0.01 + 0.2x + 0.2x + 4x^2) &= 1.7\times 10^{-5}\\ (x)(4x^2 + 0.4x + 0.01) &= 1.7\times 10^{-5}\\ 4x^3 + 0.4x^2 + 0.01x &= 1.7\times 10^{-5}\\ 4x^3 + 0.4x^2 + 0.01x - 1.7\times 10^{-5} &= 0\\ x &= 1.596 \times 10^{-3}\\[1.5ex] [\mathrm{Pb^{2+}}]_{\mathrm{eq}} &= 1.596 \times 10^{-3}~M \\ [\mathrm{Cl^-}]_{\mathrm{eq}} &= 0.1 + 2(1.596 \times 10^{-3}~M) = 0.103~M \end{align*}\]
Determine the molar concentration of the reactant by using stoichiometry and a product that is not a common ion.
\[\begin{align*} [\mathrm{PbCl_2}] &= \dfrac{1.596 \times 10^{-3}~\mathrm{mol~Pb^{2+}}}{\mathrm{L}} \left ( \dfrac{1~\mathrm{mol~PbCl_2}}{1~\mathrm{mol~Pb^{2+}}} \right ) \\[1.5ex] &= 1.596 \times 10^{-3}~M \end{align*}\]
Therefore, this saturated solution is a 1.596 × 10–3 M PbCl2 aqueous solution.
To determine the mass concentration (in g L–1), multiply the molar concentration by the molar mass of the compound.
\[\begin{align*} \rho_{\mathrm{PbCl_2}} &= \left ( 1.596 \times 10^{-3}~\mathrm{mol~L^{-1}} \right ) \left ( 278.1~\mathrm{g~mol^{-1}}\right ) \\[1.5ex] &= 0.444~\mathrm{g~L^{-3}} \end{align*}\]
Therefore, this saturated solution is a 0.444 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.
\[\begin{align*} m_{\mathrm{PbCl_2}} &= \left ( 0.444~\mathrm{g~L^{-3}} \right ) \left ( 2.0~\mathrm{L} \right ) \\ &= 0.888~\mathrm{g} \end{align*}\]
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 0.1 M Na2SO4 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 |
0.1 | |||
C |
+2x |
x | |||
E |
2x |
0.1+x |
Write out the equilibrium expression and then solve for x.
Note: In this example, we end up with a cubic function. See the calculation performed using WolframAlpha.
\[\begin{align*} [\mathrm{Ag^{+}}]^2[\mathrm{SO_4^{2-}}] &= K_{\mathrm{sp}}\\ (2x)^2(0.1+x) &= 1.2\times 10^{-5}\\ 4x^2(0.1+x) &= 1.2\times 10^{-5}\\ 0.4x^2 + 4x^3 &= 1.2\times 10^{-5}\\ 4x^3 + 0.4x^2 - 1.2\times 10^{-5} &= 0 \\ x &= 5.34\times 10^{-3}\\[2ex] [\mathrm{Ag^{+}}]_{\mathrm{eq}} &= 2(5.34\times 10^{-3}~M) = 1.068\times 10^{-2}~M\\ [\mathrm{SO_4^2-}]_{\mathrm{eq}} &= 0.1 + 5.34\times 10^{-3} = 0.105~M \end{align*}\]
Determine the molar concentration of the reactant by using stoichiometry and a product that is not a common ion.
\[\begin{align*} [\mathrm{Ag_2SO_4}] &= \dfrac{1.068\times 10^{-2}~\mathrm{mol~Ag^{2+}}}{\mathrm{L}} \left ( \dfrac{1~\mathrm{mol~Ag_2SO_4}}{2~\mathrm{mol~Ag^{2+}}} \right ) \\[1.5ex] &= 5.34\times 10^{-3}~M \end{align*}\]
Therefore, this saturated solution is a 5.34 × 10–3 M Ag2SO4 aqueous solution.
To determine the mass concentration (in g L–1), multiply the molar concentration by the molar mass of the compound.
\[\begin{align*} \rho_{\mathrm{Ag_2SO_4}} &= \left ( 5.34\times 10^{-3}~\mathrm{mol~L^{-1}} \right ) \left ( 311.799~\mathrm{g~mol^{-1}}\right ) \\[1.5ex] &= 1.66~\mathrm{g~L^{-1}} \end{align*}\]
Therefore, this saturated solution is a 1.66 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.
\[\begin{align*} m_{\mathrm{Ag_2SO_4}} &= \left ( 1.66~\mathrm{g~L^{-1}} \right ) \left ( 0.750~\mathrm{L} \right ) \\ &= 1.25~\mathrm{g} \end{align*}\]
Summary
Substances are more soluble in pure water than solutions containing a common ion. The more common ion present in solution, the lower the solubility of the subtance that contains the common ion!
| Solid | Solution | Molar Solubility | Mass Solubility |
|---|---|---|---|
AgCl |
water |
1.33 × 10–5 |
1.91 × 10–3 |
0.1 M NaCl(aq) |
1.77 × 10–9 |
2.54 × 10–7 | |
PbCl2 |
water |
1.62 × 10–2 |
4.5 |
0.1 M NaCl(aq) |
1.596 × 10–3 |
0.444 | |
Ag2SO4 |
water |
1.44 × 10–2 |
4.5 |
0.1 M Na2SO4(aq) |
5.34 × 10–3 |
1.66 |
Practice
Rank the solubility (from greatest to least) of Ca3(PO4)2 in the following solutions:
A. water
B. 0.50 M Na3PO4
C. 1.00 M Sr3(PO4)2
D. 0.75 M K3PO4
Answer
Determine the concentrations of the common ion (PO42–).
A. water → 0 M
B. 0.50 M Na3PO4 → 0.50 M
C. 1.00 M Sr3(PO4)2 → 2.00 M
D. 0.75 M K3PO4 → 0.75 M
Rank the solubility:
Solution A > B > D > C
Formation Constant
As seen with Acidic Metal Hydrides, some substances can form complex ions when in aqueous solution. The extent at which these complex ions can form is described with a new equilibrium constant called a formation constant written as Kf. Formation constants tend to be very large.
Silver(I) diammine cation, Ag(NH3)2+
When silver(I) chloride dissolves in an aqueous solution of ammonia (a basic solution), the following reaction occurs.
\[\mathrm{AgCl}(s) + \mathrm{NH_3}(aq) \rightleftharpoons \mathrm{Ag(NH_3)_2^+}(aq) + \mathrm{Cl^-}(aq) \quad K_{\mathrm{f}} = 1.7\times 10^{7}\]
The dissolved silver readily forms a complex ion with the ammonia molecules in solution (the equilibrium constant is very large!). The presence of ammonia leads to the very favored formation of the silver(I) diammine complex ion and therefore increases the solubility of AgCl(s)!
\[\begin{align*} \mathrm{AgCl}(s) &\rightleftharpoons \mathrm{Ag^+}(aq) + \mathrm{Cl^-}(aq) \quad &&K_{\mathrm{sp}} = 1.77\times 10^{-10} \\[1.5ex] \mathrm{Ag^+}(aq) + \mathrm{NH_3}(aq) &\rightleftharpoons \mathrm{Ag(NH_3)_2^+}(aq) \quad &&K_{\mathrm{f}} = 1.7\times 10^{7} \end{align*}\]
As AgCl dissociates, Ag+(aq) concentration is consumed as it forms a complex ion with ammonia. According to Le Chatelier’s principle, the removal of product in the first reaction (due to the second reaction) will drive the first reaction to the right to produce more product (thereby dissolving more AgCl)!