Question

The solubility of a certain sparingly soluble substance $M{X}_n$ is nearly $1.4\times {10}^{-4}M$. If the solubility product is $1.1\times {10}^{-11}$, what is the value of $n$?

• A. 1
• B. 2
• C. 3
• D. 1.5

Answer 1

The correct option is B 2

As given,

$M{X}_n\to M^{+n}+n{X}^{-}$

$S$ $nS$

S = $1.4\times {10}^{-4}$

Solubility product $=S\times (nS{)}^n=n^n{S}^{n+1}=1.1\times {10}^{-11}$

By comparison, $n=2.$