Question

. Isotonic saline solution is 0.154 M NaCl(aq) What is the solubility of AgCl (Ksp = 1.8 × 10-10 ) in such a solution?

a. 2.8 × 10-11 M

b. 1.2 × 10-9 M

c. 5.3 × 10-6 M

d. 3.4 × 10-5 M

e. 0.077 M

Answer 1

AgCl (s) < => Ag+ (aq) + Cl- (aq)

for this the equilibrium condition required that

Ksp = 1.80 x 10^-10 = [Ag+][Cl-]

let x = mol/L AgCl that dissolves.
This will put into the solution x mol/L Ag+ and x mol/L CL-
But the solution arleady contains 0.154 M NaCl.
Since NaCl is a strong electrolyte 0.154 M NaCl corresponds to 0.154 M Cl-
At equilibrium :
[Ag+]= x and [Cl-]= x + 0.154

1.80 x 10^-10 = x ( x + 0.154)

=>x^2+0.154x-1.8*10^-10=0

x = molar solubility = 1.168 x 10^-9

it is very close to option b hence option b is the correct answer