Question

The relationship between $K_{sp}$ and molar solubility $\left(x\right)$ of AgCl in 0.2M KCl is:

• A. $K_{\left(sp\right)}=x$
• B. $K_{\left(sp\right)}=\frac{x}{0.2}$
• C. $K_{\left(sp\right)}=0.2x$
• D. $K_{\left(sp\right)}=x^{\frac{1}{2}}$

Answer 1

Answer: (C)

$K_{\left(sp\right)}=0.2x$

$\underset{x}{AgCl}\rightleftharpoons \underset{x}{A{g}^{+}}+\underset{0.2+x}{C{l}^{-}}$
The expression for the solubility product is $K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]=x(x+0.2)$.
But, $x<<<0.2$,
thus, $K_{sp}=0.2x$.