Calculate the solubility of silver chloride (in g/L) in a $6.5 × 10^{-3}$ M silver nitrate solution.

Strategy This is a common-ion problem. The common ion here is $Ag^{+}$ , which is supplied by both AgCl and $AgNO_{3}$ . Remember that the presence of the common ion will affect only the solubility of AgCl (in g/L), but not the $K_{sp}$ value because it is an equilibrium constant.

Question

Calculate the solubility of silver chloride (in g/L) in a [latex]6.5 × 10^{-3}[/latex] M silver nitrate solution.

Strategy This is a common-ion problem. The common ion here is [latex]Ag^{+}[/latex] , which is supplied by both AgCl and [latex]AgNO_{3}[/latex] . Remember that the presence of the common ion will affect only the solubility of AgCl (in g/L), but not the [latex]K_{sp}[/latex] value because it is an equilibrium constant.

Accepted Answer

Step 1: The relevant species in solution are [latex]Ag^{+}[/latex] ions (from both AgCl and [latex]AgNO_{3}[/latex]) and [latex]Cl^{-}[/latex] ions. The ions are [latex]NO_3^-[/latex] spectator ions.

Step 2: Because [latex]AgNO_{3}[/latex] is a soluble strong electrolyte, it dissociates completely:

[latex]\begin{matrix} AgNO_{3} & \overset{H_{2}O }{ ightarrow }&Ag^{+}(aq)&+&NO^{-}_{3} (aq) \ &&6.5 imes 10^{-3} M&&6.5 imes 10^{-3} M \end{matrix} [/latex]

Let s be the molar solubility of AgCl in [latex]AgNO_{3}[/latex] solution. We summarize the changes in concentrations as follows:

[latex]\begin{matrix} & AgCl(s) & & \xrightleftharpoons[ ]{} & Ag^{+} (aq)&+&Cl^{-} (aq) \ Initial (M): & & &&6.5 imes 10^{-3} &&0.00 &&&&& \ Change (M): & -s & &&+s &&+s &&&& \ \hline Equilibrium (M):&&&&(6.5 imes 10^{-3}+s)&&s && \end{matrix}[/latex]

Step 3:

[latex]K_{sp}=[Ag^{+}][Cl^{-}][/latex]

[latex]1.6 imes 10^{-10} =(6.5 imes 10^{-3}+s )(s)[/latex]

Because AgCl is quite insoluble and the presence of [latex]Ag^{+}[/latex] ions from [latex]AgNO_{3}[/latex] further lowers the solubility of AgCl, s must be very small compared with [latex]6.5 × 10^{-3}[/latex]. Therefore, applying the approximation [latex]6.5 × 10^{-3} + s ≈ 6.5 × 10^{-3}[/latex] , we obtain

[latex]1.6 imes 10^{-10} =(6.5 imes 10^{-3})s[/latex]

[latex]s=2.5 imes 10^{-8} M[/latex]

Step 4: At equilibrium
[latex][Ag^{+}]=(6.5 imes 10^{-3}+2.5 imes 10^{-8} ) M \approx 6.5 imes 10^{-3} M[/latex]

[latex][Cl^{-}]=2.5 imes 10^{-8} M[/latex]

and so our approximation was justified in step 3. Because all the [latex]Cl^{-}[/latex] ions must come from AgCl, the amount of AgCl dissolved in [latex]AgNO_{3}[/latex] solution also is [latex]2.5 imes 10^{-8} M[/latex]. Then, knowing the molar mass of AgCl (143.4 g), we can calculate the solubility of AgCl as follows:

solubility of AgCl in [latex]AgNO_{3}[/latex] solution=[latex]\frac{2.5 imes 10^{-8} mol AgCl}{1 L soln} imes \frac{143.4 g AgCl}{1 mol AgCl} [/latex]

[latex]=3.6 imes 10^{-6} g/L[/latex]

Check The solubility of AgCl in pure water is [latex]1.9 imes 10^{-3} g/L[/latex] (see the Practice Exercise in Example 16.9). Therefore, the lower solubility ([latex]3.6 imes 10^{-6} g/L[/latex]) in the presence of [latex]AgNO_{3}[/latex] is reasonable. You should also be able to predict the lower solubility using Le Châtelier’s principle. Adding [latex]Ag^{+}[/latex] ions shifts the equilibrium to the left, thus decreasing the solubility of AgCl.

Question 16.12: Calculate the solubility of silver chloride (in g/L) in a 6....

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