6.2 Common Ion
Solutions already containing an ion that is in common with the substance being dissolved naturally lowers the solubility of the starting substance.
Figure 6.1: Solubility of some solids in water containing a common ion
6.2.1 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 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.
\[\begin{align*} m_{\mathrm{AgCl}} &= \left (2.54\times 10^{-7}~\mathrm{g~L^{-1}} \right ) \left ( 5.0~\mathrm{L} \right ) \\ &= 1.27\times 10^{-6}~\mathrm{g} \end{align*}\]
6.2.2 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 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.
Cubic Approach
We can end up with a cubic function. See the calculation performed using WolframAlpha.
\[\begin{align*} [\mathrm{Pb^{2+}}][\mathrm{Cl^-}]^2 &= K_{\mathrm{sp}}\\ (x)(0.1 + 2x)^2 &= 1.7\times 10^{-5}\\ (x)(0.01 + 0.2x + 0.2x + 4x^2) &= 1.7\times 10^{-5}\\ (x)(4x^2 + 0.4x + 0.01) &= 1.7\times 10^{-5}\\ 4x^3 + 0.4x^2 + 0.01x &= 1.7\times 10^{-5}\\ 4x^3 + 0.4x^2 + 0.01x - 1.7\times 10^{-5} &= 0\\ x &= 1.596 \times 10^{-3}\\[1.5ex] [\mathrm{Pb^{2+}}]_{\mathrm{eq}} &= 1.596 \times 10^{-3}~M \\ [\mathrm{Cl^-}]_{\mathrm{eq}} &= 0.1 + 2(1.596 \times 10^{-3}~M) = 0.103~M \end{align*}\]
Small x Approach
Using the small x approximation, we get
\[\begin{align*} [\mathrm{Pb^{2+}}][\mathrm{Cl^-}]^2 &= K_{\mathrm{sp}}\\ (x)(0.1 + 2x)^2 &= 1.7\times 10^{-5}\\ (x)(0.1)^2 &= 1.7\times 10^{-5}\\ x &= 1.75 \times 10^{-3} \quad (3.5\%) \\[1.5ex] [\mathrm{Pb^{2+}}]_{\mathrm{eq}} &= 1.75 \times 10^{-3}~M \\ [\mathrm{Cl^-}]_{\mathrm{eq}} &= 0.1 + 2(1.75 \times 10^{-3}~M) = 0.104~M \end{align*}\]
Note: I will use the value for x determined from the cubic approach but note that I could use the x from the small x approach and get very similar answers.
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.
\[\begin{align*} \rho_{\mathrm{PbCl_2}} &= \left ( 1.596 \times 10^{-3}~\mathrm{mol~L^{-1}} \right ) \left ( 278.1~\mathrm{g~mol^{-1}}\right ) \\[1.5ex] &= 0.444~\mathrm{g~L^{-1}} \end{align*}\]
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.
\[\begin{align*} m_{\mathrm{PbCl_2}} &= \left ( 0.444~\mathrm{g~L^{-1}} \right ) \left ( 2.0~\mathrm{L} \right ) \\ &= 0.888~\mathrm{g} \end{align*}\]
6.2.3 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 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.
Cubic Approach
We can end up with a cubic function. See the calculation performed using WolframAlpha.
\[\begin{align*} [\mathrm{Ag^{+}}]^2[\mathrm{SO_4^{2-}}] &= K_{\mathrm{sp}}\\ (2x)^2(0.1+x) &= 1.2\times 10^{-5}\\ 4x^2(0.1+x) &= 1.2\times 10^{-5}\\ 0.4x^2 + 4x^3 &= 1.2\times 10^{-5}\\ 4x^3 + 0.4x^2 - 1.2\times 10^{-5} &= 0 \\ x &= 5.34\times 10^{-3}\\[2ex] [\mathrm{Ag^{+}}]_{\mathrm{eq}} &= 2(5.34\times 10^{-3}~M) = 1.068\times 10^{-2}~M\\ [\mathrm{SO_4^2-}]_{\mathrm{eq}} &= 0.1 + 5.34\times 10^{-3} = 0.105~M \end{align*}\]
Small x approach
Unfortunately, the small x approximation leads to an x that is a bit too large (5.5%).
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.
\[\begin{align*} m_{\mathrm{Ag_2SO_4}} &= \left ( 1.66~\mathrm{g~L^{-1}} \right ) \left ( 0.750~\mathrm{L} \right ) \\ &= 1.25~\mathrm{g} \end{align*}\]
6.2.4 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 liquids and/or aqueous solutions:
A. pure water
B. 0.50 M Na3PO4
C. 1.00 M Sr3(PO4)2
D. 0.75 M K3PO4
Solution
Determine the concentrations of the common ion (PO42–).
A. pure 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