Calculate the molar solubility of AgBr in:

Question

a. 0.0300 M AgNO₃ solution

b. 0.100 M NaBr solution

Answer 1

AgBr(s) <==> Ag⁺(aq) + Br^-(aq) ... Ksp = 5.0x10^-13

(a). 0.0300 M AgNO₃. Note that Ag⁺ is the common ion.

Ksp = [Ag⁺][Br-]

5.0x10^-13 = [0.0300][Br-]

[Br- ] = 1.67x10^-11 M = solubility of AgCl in the presence of 0.03 M AgNO₃

(b). 0.100 M NaBr. Note that Br- is the common ion.

Ksp = [Ag⁺][Br-]

5.0x10^-13 = [Ag⁺][0.100]

[Ag⁺] = 5.00x10^-12 M = solubility of AgCl in the presence of 0.10 M NaBr

As a note of interest, let's look at the solubility of AgCl in the absence of any common ion:

5.0x10^-13 = [Ag⁺][Cl-]

5.0x10-13 = (x)(x)

x = 7.1x10^-7 M = solubility of AgCl in the absence of any common ion. Note it is greater than when a common ion is present. This is in agreement with Le Chatelier's principle.