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

Answer:

$2.0x10^{-5}\frac{mol}{L}$

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:

$2.0x10^{-5}\frac{mol}{L}$

Best regards!