Question

19.62 The molar solubility of ${\mathbf{\text{M}}}_{\mathbf{2}}\mathbf{\text{X}}$is$\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathbf{\text{M}}$. What is the molarity of each ion? How do you set up the calculation to find${\mathbf{K}}_{\mathbf{\text{sp}}}$? What assumption must you make about the dissociation of${\mathbf{\text{M}}}_{\mathbf{2}}\mathbf{\text{X}}$into ions? Why is the calculated${\mathbf{K}}_{\mathbf{\text{sp}}}$higher than the actual value?

Short answer

We must assume that ${\mathrm{M}}_2\mathrm{X}$is partially dissolved into its ions and that there are no additional sources of ${\mathrm{M}}^{+}$and role="math" localid="1663408436372" ${\mathrm{X}}^2$ions in the solution.

If there are other ${\mathrm{M}}^{\text{'}}$ and ${\mathrm{X}}^2$ ions from other sources in the solution, the computed Ksp value may be higher than the real value.

Step-by-step solution

Definition of pH

Solution's pHvalue, which measures the concentration of hydrogen ions, reveals whether it is acidic or alkaline.

Find the molarities

To answer this problem, we must first write the dissolution equation for the following: ${\text{M}}_2\mathrm{X}$:

$$\begin{gathered} {\mathrm{M}}_2\mathrm{X}\left(\mathrm{s}\right)>>2{\mathrm{M}}^{\perp }\left(\mathrm{aq}\right)+{\mathrm{X}}^2\left(\mathrm{aq}\right) \\ \mathrm{Ksp}={\left[{\mathrm{M}}^{†}\right]}^2\left[{\mathrm{X}}^2\right] \end{gathered}$$

We squared the $\left[{\text{M}}^{-}\right]$because there is coefficient 2 in the dissolution equation.

The molarities of ${\mathrm{M}}^{+}$and ${\mathrm{X}}^2$may now be determined. Because we acquire two mols of ions when one mol of solid compound dissolves, the molarity of ${\mathrm{M}}^{+}$will be equal to the twofold value of solubility that we are given $(\mathrm{S}=5\times {10}^{-5}\mathrm{M})$

$$\begin{gathered} \left[{\mathrm{M}}^{†}\right]=2\mathrm{S}=1\times {10}^{-4}\mathrm{Ml} \\ \left[{\mathrm{X}}^2\right]=\mathrm{S}=5\times {10}^5\mathrm{M} \end{gathered}$$

Therefore, the molarities is:$$\begin{gathered} \left[{\mathrm{M}}^{†}\right]=2\mathrm{S}=1\times {10}^{-4}\mathrm{M} \\ \left[{\mathrm{X}}^2\right]=\mathrm{S}=5\times {10}^5\mathrm{Md} \end{gathered}$$

Calculate the Ksp

Now, we can calculate the Ksp:

$$\begin{gathered} \mathrm{Ksp}={\left[{\mathrm{M}}^{+}\right]}^2 \\ \left[{\mathrm{X}}^2\right]=(2\mathrm{S}{)}^2 \\ \mathrm{S}=4 \\ {\mathrm{S}}^3=5\times {10}^{13} \end{gathered}$$

When performing these calculations, we must assume that ${\mathrm{M}}_2\mathrm{X}$is partially dissolved into its ions and that there are no other sources of $\mathrm{M}$and ${\mathrm{X}}^2$ions in the solution.

As a result, if other M and ${\mathrm{X}}^2$ions from other sources are present in the solution, the computed Ksp value may be larger than the real value.