Question

You add an excess of solid MX in 250 g of water. You measure the freezing point and find it to be $-0.028°C$. What is the $K_{sp}$ of the solid?

Short answer

The $K_{sp}$ of the solid is$K_{sp}=5.67\times {10}^{-5}mo{l}^2/L^2$

Step-by-step solution

Step 1:

MX dissociates into one ions and two ions. So, Van’t Hoff factor is 2.

Calculate the Molarity for freezing point depression.

$$\begin{gathered} \Delta T=i\times K_f\times m \\ m=\frac{\Delta T}{i\times K_f}=\frac{0.028°C}{2\times 1.86°Ckgmo{l}^{-1}} \\ m=7.528\times {10}^{-3}mol/L \end{gathered}$$

Step 2:

Assume that MX salt completely dissociates, therefore the equilibrium concentrations are as $\left[M^{+}\right]=\left[X^{-}\right]=M=7.527\times {10}^{3}mol/L$

Step 3:

Calculate the $K_{sp}$ .

$$\begin{gathered} K_{sp}=\left[M^{+}\right]\left[X^{-}\right] \\ =70527\times {10}^{-3}mol/L\times 7.527\times {10}^{-3}mol/l \\ =5.67\times {10}^{-5}mo{l}^2/L^2 \end{gathered}$$