Question

A solution saturated with a salt of the type M3X2 hasan osmotic pressure of 2.64 3 1022 atm at 258C. Calculate the K_sp value for the salt, assuming ideal behavior.

Short answer

The K_sp value for the salt is$K_{sp}=5.07\times {10}^{-17}mo{l}^2/L^2$

Step-by-step solution

Step 1:

$M_3{X}_2$ dissociates into three $M^{2+}$ions and two $X^{3-}$ions. So, Van’t Hoff factor is 5.

Calculate the Molarity for osmotic pressure.

$$\begin{gathered} \pi =M\times R\times T \\ M=\frac{\pi }{i\times R\times T}=\frac{2.64\times {10}^{-2}atm}{5\times 0.08206Latm{K}^{-1}mo{l}^{-1}\times 298K} \\ M=2.159\times {10}^{-4}mol/L \end{gathered}$$

Step 2:

Assume that $M_3{X}_2$salt completely dissociates, therefore the equilibrium concentrations are as

$$\begin{gathered} \left[M^{2+}\right]=3\times M=3\times 2.159\times {10}^{-4}mol/L=6.477\times {10}^{-4}mol/L \\ \left[X^{3-}\right]=s\times M=s\times 2.159\times {10}^{-4}mol/L=4.318\times {10}^{-4}mol/L \end{gathered}$$

Step 3:

Calculate the $K_{sp}$.

$$\begin{gathered} K_{sp}=\left[M^{2+}\right]\left[X^{3-}\right] \\ =6.477\times {10}^{-4}mol/L\times 4.318\times {10}^{-4}mol/l \\ =5.07\times {10}^{-17}mo{l}^2/L^2 \end{gathered}$$