Question
The concentration of $$A{g^ + }$$ in a saturated solution of $$A{g_{_2}}Cr{O_4}$$ at $$20{\,^ \circ }C$$ is $$1.5 \times {10^{ - 4}}\,mol\,{L^{ - 1}}.$$ The solubility product of $$A{g_{_2}}Cr{O_4}$$ at $$20{\,^ \circ }C$$ is
A.
$$1.687 \times {10^{ - 12}}$$
B.
$$1.75 \times {10^{ - 10}}$$
C.
$$3.0 \times {10^{ - 8}}$$
D.
$$4.5 \times {10^{ - 10}}$$
Answer :
$$1.687 \times {10^{ - 12}}$$
Solution :
$$A{g_{_2}}Cr{O_4} \rightleftharpoons 2A{g^ + } + CrO_4^{2 - }$$
$$A{g^ + } = 1.5 \times {10^{ - 4}}\,M;$$ $$CrO_4^{2 - } = 0.75 \times {10^{ - 4}}\,M$$ $$\left( {\frac{1}{2}\,{\text{of}}\,\,A{g^ + }} \right)$$
$$\eqalign{
& {K_{sp}} = {\left( {1.5 \times {{10}^{ - 4}}} \right)^2} \times \left( {0.75 \times {{10}^{ - 4}}} \right) \cr
& \,\,\,\,\,\,\,\,\,\, = 1.687 \times {10^{ - 12}} \cr} $$