Question

Manganese (II) sulfide is one of the compounds found in the nodules on the ocean floor that may eventually be a primary source of many transition metals. The solubility of $\mathbf{MnS}\mathbf{is}\mathbf{4}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{100}\mathbf{}\mathbf{mL}$ solution. Estimate the K_sp of $\mathbf{\text{MnS}}$.

Short answer

The value of ${\mathrm{K}}_{\mathrm{sp}}=2.9\times {10}^{-9}.$

Step-by-step solution

Concept introduction

Manganese (II) Sulfide is a manganese-sulfur chemical compound. Alabandite, rambergite, and the recently discovered browneite are all minerals that include it.

Estimation of  Ksp

The dissolving equation for$\text{MnS}$can be written as follows

role="math" localid="1663676361687" $$\begin{gathered} \mathrm{MnS}\left(\mathrm{s}\right)\rightleftharpoons {\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{S}}^{2-}\left(\mathrm{aq}\right) \\ {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Mn}}^{2+}\right]\left[{\mathrm{S}}^{2-}\right] \\ {\mathrm{K}}_{\mathrm{sp}}={\mathrm{x}}^2 \end{gathered}$$

We can write down the concentration of the ions in terms of moles per liter.

$$\begin{gathered} \mathrm{x}=\frac{4.7\times {10}^{-4}\mathrm{g}}{100\mathrm{ml}}\times \frac{1\mathrm{mol}}{87.003\mathrm{g}}\times \frac{1000\mathrm{ml}}{1\mathrm{l}} \\ \mathrm{x}=5.4\times {10}^{-5}\mathrm{M} \\ {\mathrm{K}}_{\mathrm{sp}}={\left(5.4\times {10}^{-5}\mathrm{M}\right)}^2 \\ {\mathrm{K}}_{\mathrm{sp}}=2.9\times {10}^{-9} \end{gathered}$$

Therefore, the Estimation of ${\mathrm{K}}_{\mathrm{sp}}=2.9\times {10}^{-9}$.