Q19.62P - Chemistry: Molecular Nature Of Matter And Change

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?

Step-by-Step Solution

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 ${\mathrm{X}}^2$ 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 1: Definition of pH

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

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

Step 3: 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.