Question

The K_sp of AgCl is 1.6 times 10^10. What is the solubility of AgCl in 0.0010 M KCl? Give your answer using scientific notation and to 2 significant figures(i.e., one decimal place).

Answer 1

Answer:

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

for which the equilibrium condition required that

Ksp = 1.60 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 0.001 M KCl.
Since KCl is a strong electrolyte 0.001 M NaCl corresponds to 0.001 M Cl-
At equilibrium :
[Ag+]= x and [Cl-]= x + 0.001

1.60 x 10^-10 = x ( x + 0.001)

x = molar solubility = 1.599 x 10^-7 M

The solubility of AgCl=1.599 x 10^-7 M