If the solubility of $\mathrm{MX}_2$ type electrolyte is $0.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$…

If the solubility of $\mathrm{MX}_2$ type electrolyte is $0.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$, then, find out $\mathrm{K}_{s p}$ of the electrolyte.
  1. $5 \times 10^{-12}$
  2. $25 \times 10^{-10}$
  3. $1 \times 10^{-13}$
  4. $5 \times 10^{-13}$

Solution

An electrolyte \(\mathrm{MX}_2\) undergoes dissociation as follows :-
\(\mathrm{MX}_2 \rightleftharpoons \mathrm{M}^{+2}+2 \mathrm{X}^{-}\)
\(\begin{array}{|c|c|c|c|}
\hline \text {Concentration } & \mathrm{MX}_2 & \mathrm{M}^{+2} & \mathrm{X}^{-} \\
\hline \text {Initial concentration } & 1 & 0 & 0 \\
\hline \text {Concentration at Equilibrium } & 1-\mathrm{s} & \mathrm{s} & 2 \mathrm{~s} \\
\hline
\end{array}\)
Thus from the above condition we can say that,
\(\mathrm{K}_{\mathrm{sp}}=\mathrm{s} \times(2 \mathrm{~s})^2=4 \times(\mathrm{s})^3\)
Here, s (the solubility) is \(0.5 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}\).
\(\begin{aligned}
& \therefore \mathrm{K}_{\mathrm{sp}}=4 \times\left(0.5 \times 10^{-4}\right)^3 \\
& \therefore \mathrm{K}_{\mathrm{sp}}=5 \times 10^{-13}
\end{aligned}\)

Asked in: NEET 2002

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