Question
What will be the solubility of $$AgCl$$ in $$0.05\,M\,NaCl$$ aqueous solution if solubility product of $$AgCl$$ is $$1.5 \times {10^{ - 10}}?$$
A.
$$3 \times {10^{ - 9}}\,mol\,{L^{ - 1}}$$
B.
$$0.05\,mol\,{L^{ - 1}}$$
C.
$$1.5 \times {10^{ - 5}}\,mol\,{L^{ - 1}}$$
D.
$$3 \times {10^9}\,mol\,{L^{ - 1}}$$
Answer :
$$3 \times {10^{ - 9}}\,mol\,{L^{ - 1}}$$
Solution :
$$\eqalign{
& {K_{sp}} = \left[ {A{g^ + }} \right]\left[ {C{l^ - }} \right] \cr
& \left[ {C{l^ - }} \right] = NaCl = 0.05\,M \cr
& A{g^ + } = \frac{{1.5 \times {{10}^{ - 10}}}}{{0.05}} = 3 \times {10^{ - 9}}\,M; \cr
& \left[ {A{g^ + }} \right] = {\text{solubility}} = 3 \times {10^{ - 9}}\,M \cr} $$