Question

Determine the molar solubility of CuCl in a solution containing 0.050 KCl. Ksp of CuCl is 1.0 x 10-6

Answer 1

`2.0x10^(-5)(mol)/(L)`

Step-by-step explanation:

Hello!

In this case, since the dissolution of copper (I) chloride is:

`CuCl(s)\rightarrow Cu^++Cl^-`

And its equilibrium expression is:

`Ksp=[Cu^+][Cl^-]`

We can represent the molar solubility via the reaction extent as
`x`, however, since there is 0.050 M KCl we immediately add such amount to the chloride ion concentration since KCl is readily ionized; therefore we write:

`1.0x10^(-6)=(x)(0.050+x)`

Thus, solving for
`x`, we obtain:

`1.0x10^(-6)=0.050x+x^2\\\\x^2+0.050x-1x10^(-6)=0`

By using the quadratic equation, we obtain:

`x_1=2.0x10^(-5)M\\\\x_2=-0.05M`

Clearly, the solution is
`x_1=2.0x10^(-5)M` because no negative results are

allowed. Therefore, the molar solubility is:

`2.0x10^(-5)(mol)/(L)`

Best regards!