Question

For a sparingly soluble salt $A_p{B}_q$, the relationship of its solubility product $\left(L_S\right)$ with its solubility
($S$) is:

• A. $L_S=S^{p+q}\cdot p^p\cdot q^q$
• B. $L_S=S^{p+q}\cdot p^q\cdot q^p$
• C. $L_S=S^{pq}\cdot p^p\cdot q^q$
• D. $L_S=S^{pq}\cdot (pq{)}^{p+q}$

Answer 1

Answer: (A)

$L_S=S^{p+q}\cdot p^p\cdot q^q$

The dissociation equilibrium of sparingly soluble salt can be represented as
$A_p{B}_q\rightleftharpoons p{A}_{ps}^{+}+q{B}_{qs}^{-}$

Let $S$ be the solubility in mol/litre. The expression for the solubility product will be
$K_{sp}=\left[A^{+}{]}^p\right[B^{-}{]}^q=\left(pS{)}^p\right(qS{)}^q=p^p{q}^q(S{)}^{p+q}$.