A solution contains 0.020 M Cl^− ions and 0.020 M Br^− ions. To separate the Cl^− ions from the Br^− ions, solid AgNO3 is slowly added to the solution without changing the volume. What concentration of Ag^+ ions (in mol/L) is needed to precipitate as much AgBr as possible without precipitating AgCl

Question

A solution contains 0.020 M [latex]Cl^{-}[/latex] ions and 0.020 M [latex]Br^{-}[/latex] ions. To separate the [latex]Cl^{-}[/latex] ions from the [latex]Br^{-}[/latex] ions, solid [latex]AgNO_{3}[/latex] is slowly added to the solution without changing the volume. What concentration of [latex]Ag^{+}[/latex] ions (in mol/L) is needed to precipitate as much AgBr as possible without precipitating AgCl?

Strategy In solution, [latex]AgNO_{3}[/latex] dissociates into [latex]Ag^{+}[/latex] and ions. The [latex]Ag^{+}[/latex] ions then combine with the [latex]Cl^{-}[/latex]and [latex]Br^{-}[/latex] ions to form AgCl and AgBr precipitates. Because AgBr is less soluble (it has a smaller [latex]K_{sp}[/latex] than that of AgCl), it will precipitate first. Therefore, this is a fractional precipitation problem. Knowing the concentrations of [latex]Cl^{-}[/latex] and [latex]Br^{-}[/latex] ions, we can calculate [latex][Ag^{+}][/latex] from the [latex]K_{sp}[/latex] values. Keep in mind that [latex]K_{sp}[/latex] refers to a saturated solution. To initiate precipitation, [latex][Ag^{+}][/latex] must exceed the concentration in the saturated solution in each case.

Accepted Answer

The solubility equilibrium for AgBr is

[latex]AgBr(s) \xrightleftharpoons[]{} Ag^{+}(aq) + Br^{-}(aq) K_{sp} = [Ag^{+}][Br^{-}][/latex]

Because [latex][Br^{-}][/latex] = 0.020 M, the concentration of [latex]Ag^{+}[/latex] that must be exceeded to initiate the precipitation of AgBr is

[latex][Ag^{+}]=\frac{K_{sp}}{[Br^{-}]}=\frac{7.7 imes10 ^{-13} }{0.020 }[/latex]

[latex]=3.9 imes10 ^{-11} M[/latex]

Thus, [latex][Ag^{+}] >3.9 imes10 ^{-11} M[/latex] is required to start the precipitation of AgBr. The solubility equilibrium for AgCl is

[latex]AgCl(s) \xrightleftharpoons[]{} Ag^{+}(aq) + Cl^{-}(aq) K_{sp} = [Ag^{+}][Cl^{-}][/latex]

so that

[latex][Ag^{+}]=\frac{K_{sp}}{[Cl^{-}]}=\frac{1.6 imes10 ^{-10} }{0.020} [/latex]

[latex]=8.0 imes10 ^{-9} M[/latex]

Therefore, [latex][Ag^{+}] >8.0 imes10 ^{-9} M[/latex] is needed to initiate the precipitation of AgCl. To precipitate the [latex]Br^{-}[/latex] ions as AgBr without precipitating the [latex]Cl^{-}[/latex] ions as AgCl, then, [latex][Ag^{+}][/latex] must be greater than [latex]3.9 imes 10 ^{-11} M[/latex] and lower than [latex]8.0 imes10 ^{-9} M[/latex] .

Question 16.11: A solution contains 0.020 M Cl^− ions and 0.020 M Br^− ions....

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