Question

Ag₂S(s)has a larger molar solubility than CuS eventhough Ag₂S has the smaller K_sp value. Explain how this is possible.

Short answer

The greater molar solubility of Ag₂S compared to CuS can be explained based on their K_sp values.

Step-by-step solution

Solubility and solubility product

Ag₂S has higher molar solubility than CuS and the smallest K_sp.

The maximum amount of substances that dissolve a particular solution is called solubility. The equilibrium constant of molar concentration for compounds that raises to power is called the solubility product. It can be explained by:

$$\begin{gathered} A_m{B}_n\left(s\right)\rightleftharpoons m{A}^{n+}\left(aq\right)+n{B}^{m-}\left(aq\right) \\ CuS\left(s\right)\rightleftharpoons C{u}^{2+}\left(aq\right)+S^{2-}\left(aq\right) \\ {K}_{sp}CuS\left(s\right)=8.5\times {10}^{-45} \\ A{g}_{2}S\left(s\right)\rightleftharpoons 2A{g}^{+}\left(aq\right)+S^{2-}\left(aq\right) \\ {K}_{sp}A{g}_{2}S\left(s\right)\rightleftharpoons 1.6\times {10}^{-49} \end{gathered}$$

There is a production of two ions by CuS(s), whereas three ions by Ag₂S(s).

Calculating the solubility of each compound

The solubility of each compound can be calculated as:

$$\begin{gathered} CuS\left(s\right)\rightleftharpoons C{u}^{2+}\left(aq\right)+S^{2-}\left(aq\right) \\ {K}_{sp}=\left[C{u}^{2+}\left(aq\right)\right]\left[S^{2-}\left(aq\right)\right] \\ 8.5\times {10}^{-45}mo{l}^2{L}^{-2}=x^2 \\ x=9.2\times {10}^{-23}mol{L}^{-1} \\ {K}_{sp}=\left[{}_{g^{+}}{}^A[S^{2-}\right(aq) \\ 1.6\times {10}^{-49}mo{l}^2{L}^{-2}=x^3 \\ x=3.4\times {10}^{-17}mol{L}^{-1} \end{gathered}$$

Based on these calculations, it is clear that Ag₂S has a larger molar solubility than CuS .