Chemistry Solubility Math Question 13

Question

K_sp for AgCl is 1.77E-10. What is the molar solubility of AgCl in a solution of 0.28 M NaCl?

Answer 1

Unlike the previous Ksp questions posed, this one deals with the COMMON ION EFFECT. In this case, the common ion is Cl- because it is present in both AgCl and NaCl. Because the [Cl-] from the NaCl is so large compared to that from AgCl, the [Cl-] from AgCl can be ignored. Thus, we have the following situation:

AgCl(s) <==> Ag⁺(aq) + Cl-(aq)

Ksp = 1.77x10^-10 = [Ag⁺][Cl-]

We can use 0.28 M for the [Cl-] and solve for the [Ag⁺], which will give us the molar solubility under these conditions.

1.77x10^-10 = (x)(0.28)

0.28x = 1.77x10^-10

x = 6.3x10^-10 M = molar solubility of AgCl in 0.28 M NaCl