Question

For a sparingly soluble salt $A_p{B}_q$ , the relationship between its solubility product $K_{sp}$ and its solubility $\left(s\right)$ is:

• A. $K_{sp}=s^{p+q}.p^p.q^q$
• B. $K_{sp}=s^{p+q}.p^q.q^p$
• C. $K_{sp}=s^{pq}.p^p.q^q$
• D. $K_{sp}=s^{pq}.{\left(pq\right)}^{p+q}$

Answer 1

The correct option is A $K_{sp}=s^{p+q}.p^p.q^q$

The dissociation equilibrium of a sparingly soluble salt (say $A_p{B}_q$) can be represented as follows:

$A_p{B}_q\rightleftharpoons p{A}^{a+}+q{B}^{b-}$

Let $s$ be the solubility in mol/litre.

The expression for the solubility product will be as given below:

$K_{sp}=\left[A^{a+}{]}^p\right[B^{b-}{]}^q=\left(ps{)}^p\right(qs{)}^q=p^p{q}^q(s{)}^{p+q}$