Question

2. Calculate the molar solubility of the following:

a. AgCl added to pure water

b. AgCl added to water that is already 0.25 M Cl-

Answer 1

a

AgCl can dissociate in to

  AgCl ------> Ag+ + Cl-

Ksp = [Ag+][Cl-] = 1.8 x 10^-10 (this value is standadered value taken from the text book)


Let [Ag+] = s. Then [Cl-] = s, and substituting in the Ksp expression we have

(s)(s) = 1.8 x 10^-10

s² =1.8 x 10^-10

s= 1.34 x 10^-5 mol /L

= molar solubility of AgCl

Part B

Ksp = [Ag+][Cl-] = 1.8 x 10^-10 (this value is standadered value taken from the text book)

but here in the solution already 0.25 mol/L Cl^- is there

t [Ag+] = s. Then [Cl-] = s +0.25

(s)(s+0.25) = 1.8 x 10^-10

s² + 0.25s - 1.8 x 10^-10 = 0

solve the quadratic equation

s =7.20 x 10^-10 mol/L