Q19.111P - Chemistry: Molecular Nature Of Matter And Change

Question

Cadmium ion in solution is analyzed by precipitation as the sulfide, a yellow compound used as a pigment in everything from artists' oil paints to glass and rubber. Calculate the molar solubility of cadmium sulfide at $\mathbf{25}\mathbf{ }\mathbf{°}\mathbf{C}$.

Step-by-Step Solution

Answer

The molar solubility of cadmium sulfide is $1\times {10}^{-12}\text{ M}$.

Concept introduction

The number of moles of a solute that may be dissolved per litre of solution before the solution becomes saturated is known as molar solubility.

Explanation of the cadmium sulphide with table and expression

Determine the expression of the cadmium sulphide solubility product  K_sp.

 $\text{CdS(s)}\rightleftharpoons {\text{Cd}}^{\text{2 + }}\left(\text{aq}\right){\text{ + S}}^{\text{ - 2}}\left(\text{aq}\right)$

 ${\text{K}}_{\text{sp}}\text{ = }\left[{\text{Cd}}^{\text{ + 2}}\right]\left[{\text{S}}^{\text{2 - }}\right]$

The value of the solubility product  of cadmium sulphide $1\times {10}^{-24}$ at $25 °\mathrm{C}$.

Create an ICE table to represent the equilibrium. Let S represent the molar solubility.

 Now, substitute for $\left[{\text{Cd}}^{\text{ + 2}}\right]\text{ and }\left[{\text{S}}^{\text{2 - }}\right]$ and the value of K_sp

 $$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Cd}}^{+2}\right]\left[{\mathrm{S}}^{2-}\right] \\ 1\times {10}^{-24}=\mathrm{S}\times \mathrm{S} \\ \mathrm{S}=\sqrt{1\times {10}^{-24}} \\ \mathrm{S}=1\times {10}^{-12}\mathrm{M} \end{gathered}$$

Therefore, the molar solubility of cadmium sulfide is $1\times {10}^{-12}\text{M}$.