Question

The solubility of the ionic compound M₂X₃, having a molar mass of 288 g/mol, is 3.60 x 10^-7 g/L. Calculate the K_sp of the compound.

Short answer

Thevalue of K_sp of the compound with solubility$3.6\times {10}^{-7}is3.3\times {10}^{-43}$

Step-by-step solution

Conversion of solubility into mol/L

The solubility can be denoted as:

$$\begin{gathered} M_2{X}_3=\frac{3.6\times {10}^{-7}g/L}{288g/mol} \\ =1.25\times {10}^{-9}mol/L \end{gathered}$$

Calculating the value of dissolution equilibrium

The equation can be denoted as:

$$\begin{gathered} M_2{X}_3\left(s\right)\iff 2M^{3+}\left(aq\right)+3X^{-2}\left(aq\right) \\ {K}_{sp}=\left[M^{3+}\right]\left[X^{2-}\right] \\ \left[M^{3+}\right]=2smol/L \\ \left[X^{2-}\right]=3smol/L \\ {K}_{sp}={\left[2s\right]}^2{\left[3s\right]}^3=108s^5 \\ {K}_{sp}=3.3\times {10}^{-43} \end{gathered}$$

Thevalue of K_sp of the compound with solubility$3.6\times {10}^{-7}is3.3\times {10}^{-43}$