Question
Predict if there will be any precipitate by mixing $$50\,mL$$ of $$0.01\,M\,NaCl$$ and $$50\,mL$$ of $$0.01\,M\,AgN{O_3}$$ solution. The solubility product of $$AgCl$$ is $$1.5 \times {10^{ - 10}}.$$
A.
Since ionic product is greater than solubility product no precipitate will be formed.
B.
Since ionic product is lesser than solubility product, precipitation will occur.
C.
Since ionic product is greater than solubility product, precipitation will occur.
D.
Since ionic product and solubility product are same, precipitation will not occur.
Answer :
Since ionic product is greater than solubility product, precipitation will occur.
Solution :
$$\eqalign{
& NaCl + AgN{O_3} \to AgCl + NaN{O_3} \cr
& {K_{sp}} = 1.5 \times {10^{ - 10}} \cr
& \left[ {A{g^ + }} \right] = \frac{1}{2} \times {10^{ - 2}}\,M \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.5 \times {10^{ - 2}}\,M \cr
& \left[ {C{l^ - }} \right] = \frac{1}{2} \times {10^{ - 2}}\,M \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.5 \times {10^{ - 2}}M \cr
& {K_{ip}} = \left[ {A{g^ + }} \right]\left[ {C{l^ - }} \right] \cr
& = \left( {0.5 \times {{10}^{ - 2}}} \right) \times \left( {0.5 \times {{10}^{ - 2}}} \right) \cr
& = 2.5 \times {\bf{1}}{0^{ - 5}} \cr
& {K_{ip}} > {K_{sp}} \cr} $$
When $${K_{ip}} > {K_{sp}},$$ it results in precipitation.