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