Question
The solubility product of $$BaC{l_2}$$ is $$3.2 \times {10^{ - 9}}.$$ What will be its solubility in $$mol\,{L^{ - 1}}?$$
A.
$$4 \times {10^{ - 3}}$$
B.
$$3.2 \times {10^{ - 9}}$$
C.
$$1 \times {10^{ - 3}}$$
D.
$$1 \times {10^{ - 9}}$$
Answer :
$$1 \times {10^{ - 3}}$$
Solution :
$$\eqalign{
& BaC{l_2} \rightleftharpoons B{a^{2 + }} + 2C{l^ - } \cr
& {K_{sp}} = \left[ {B{a^{2 + }}} \right]{\left[ {C{l^ - }} \right]^2} \cr
& \,\,\,\,\,\,\,\,\,\, = x \times {\left( {2x} \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\, = 4{x^3} \cr} $$
$$4{x^3} = 3.2 \times {10^{ - 9}} \Rightarrow x = 9.28 \times {10^{ - 4}}$$ $$ = 0.928 \times {10^{ - 3}} \approx 1 \times {10^{ - 3}}$$