Question

Bismuth (III) sulphide is sparingly soluble in water. The solubility product $\left(K_{sp}\right)$ for the substance is given by the expression:

• A. $\left[B{i}^{3+}{]}^2\right[S^{2-}{]}^3$
• B. $\left[B{i}^{3+}{]}^2\right[S^{2-}]$
• C. $\left[B{i}^{3+}{]}^3\right[S^{2-}{]}^2$
• D. $\left[2B{i}^{3+}{]}^2\right[3S^{2-}{]}^3$

Answer 1

The correct option is A $\left[B{i}^{3+}{]}^2\right[S^{2-}{]}^3$

In solution,

$B{i}_2{S}_3\left(s\right)\rightleftharpoons 2B{i}^{3+}\left(aq\right)+3S^{2-}\left(aq\right)$

and solubility product is product of cations and anions so:

$K_{sp}=[B{i}^{3+}{]}^2\times [S^{2-}{]}^3$

Hence,option A is correct.