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 * 10^(-3) M`.

(b) The solubility of CuCl in 0.1 M NaCl is
`9.5 * 10^(-3) M`.

Step-by-step 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 * 10^(-6)`

And, equilibrium expression is as follows.

`K_(sp) = [Cu^(+)][Cl^(-)]`

`1.2 * 10^(-6) = x * x`

x =
`1.1 * 10^(-3) M`

Hence, the solubility of CuCl in pure water is
`1.1 * 10^(-3) M`.

(b) When NaCl is 0.1 M,

`CuCl(s) \rightarrow Cu^(+)(aq) + Cl^(-)(aq)`,
`K_(sp) = 1.2 * 10^(-6)`

`Cu^(+)(aq) + 2Cl^(-)(aq) \rightleftharpoons CuCl_(2)(aq)`,
`K = 8.7 * 10^(4)`

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

`K' = K_(sp) * 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' =
`(CuCl_(2))/(Cl^(-))`

0.1044 =
`(x)/(0.1 - x)`

x =
`9.5 * 10^(-3) M`

Therefore, the solubility of CuCl in 0.1 M NaCl is
`9.5 * 10^(-3) M`.