Answer:
[tex]2.0x10^{-5}\frac{mol}{L}[/tex]
Explanation:
Hello!
In this case, since the dissolution of copper (I) chloride is:
[tex]CuCl(s)\rightarrow Cu^++Cl^-[/tex]
And its equilibrium expression is:
[tex]Ksp=[Cu^+][Cl^-][/tex]
We can represent the molar solubility via the reaction extent as [tex]x[/tex], 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:
[tex]1.0x10^{-6}=(x)(0.050+x)[/tex]
Thus, solving for [tex]x[/tex], we obtain:
[tex]1.0x10^{-6}=0.050x+x^2\\\\x^2+0.050x-1x10^{-6}=0[/tex]
By using the quadratic equation, we obtain:
[tex]x_1=2.0x10^{-5}M\\\\x_2=-0.05M[/tex]
Clearly, the solution is [tex]x_1=2.0x10^{-5}M[/tex] because no negative results are
allowed. Therefore, the molar solubility is:
[tex]2.0x10^{-5}\frac{mol}{L}[/tex]
Best regards!
