Question

The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) + 2Cl- (aq) ⇄ CuCl2- (aq) K = 8.7 x 104(a) Calculate the solubility of CuCl in pure water. (Ignore CuCl2- formation for part a).(b) Calculate the solubility of CuCl in 0.100 M NaCl solution.

Answer 1

Answer: (a) The solubility of CuCl in pure water is $1.1 \times 10^{-3} M$.

(b) The solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           $CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

$K_{sp} = 1.2 \times 10^{-6}$

And, equilibrium expression is as follows.

          $K_{sp} = [Cu^{+}][Cl^{-}]$

       $1.2 \times 10^{-6} = x \times x$

             x = $1.1 \times 10^{-3} M$

Hence, the solubility of CuCl in pure water is $1.1 \times 10^{-3} M$.

(b)  When NaCl is 0.1 M,

       $CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)$,  $K_{sp} = 1.2 \times 10^{-6}$

   $Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq)$,  $K = 8.7 \times 10^{4}$

Net equation: $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

               $K' = K_{sp} \times K$

                          = 0.1044

So for, $CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)$

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = $\frac{CuCl_{2}}{Cl^{-}}$

         0.1044 = $\frac{x}{0.1 - x}$

              x = $9.5 \times 10^{-3} M$

Therefore, the solubility of CuCl in 0.1 M NaCl is $9.5 \times 10^{-3} M$.