Question

A saturated solution of AgCl is treated with solid NaCl until the [Cl−] is 0.83 M.

What will be the resulting [Ag+] in solution? Enter your answer in scientific notation.

What percent of Ag+ remains in solution at this point? %

Answer 1

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

Ksp = [ Ag+ ] [ Cl- ] = 1.77 × 10^-10 M^2

Before treating with NaCl

[ Ag+ ] [ Cl- ] = 1.77×10^-10 M^2

If put, [Ag+ ] = x and [Cl-] = x

x^2 = 1.77×10^-10M^2

x = 1.33 × 10^-5M

Therefore,

[ Ag+ ] = 1.33 ×10^-5M

After dissolving NaCl , [ Cl- ] = 0.83M

Therefore,

0.83M × [ Ag+ ] = 1.77×10^-10M^2

[ Ag+ ] = 1.77×10^-10M^2/0.83M = 2.13×10^-10M

Therefore,

% [ Ag+ ] remaining =( 2.13×10^-10/1.33×10^-5 )×100 = 0.0016%