Posted by
**Chrissy** on
.

What is the solubility of CuCl when it is in a solution of 0.35 M NaCl?(CuCl has Ksp=1.9 x 10^-7)

The Ksp is given by Ksp=[Cu+][Cl-]

if x=[Cu+]

then in our mixture we have

x moles l^-1 [Cu+]

x moles l^-1 [Cl-]

0.35 moles l^-1 [Cl-] from the NaCl

(there is also 0.35 moles l^-1 [Na+] from the NaCl but the Na+ is a spectator)

So Ksp=x(x+0.35)

if x is small wrt 0.35 then

1.9 x 10^-7=x(0.35)

hence find x