Question
Solubility of a $${M_2}S$$ type salt is $$3.5 \times {10^{ - 6}},$$ then find out its solubility product.
A.
$$1.7 \times {10^{ - 6}}$$
B.
$$1.7 \times {10^{ - 16}}$$
C.
$$1.7 \times {10^{ - 18}}$$
D.
$$1.7 \times {10^{ - 12}}$$
Answer :
$$1.7 \times {10^{ - 16}}$$
Solution :
$$\eqalign{
& {\text{Solubility of}}\,{M_2}S\,\,{\text{salt is }}3.5 \times {10^{ - 6}}M \cr
& \mathop {{M_2}S\,}\limits_{3.5 \times {{10}^{ - 6}}M} \, \rightleftharpoons \mathop {2{M^ + }}\limits_{2 \times 3.5 \times {{10}^{ - 6}}M} + \mathop {{S^{2 - }}}\limits_{3.5 \times {{10}^{ - 6}}M} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\text{on}}\,\,100\% \,\,{\text{ionisation}}} \right) \cr
& \therefore \,\,{K_{sp}}\left( {{\text{solubility product of}}\,{M_2}S} \right) \cr
& = {\left[ {{M^ + }} \right]^2}\left[ {{S^{2 - }}} \right] \cr
& = {\left( {7.0 \times {{10}^{ - 6}}} \right)^2}\left( {3.5 \times {{10}^{ - 6}}} \right) \cr
& = 171.5 \times {10^{ - 18}} \cr
& = 1.71 \times {10^{ - 16}}{\left[ M \right]^3} \cr} $$