Predict if there will be any precipitate by mixing 50 mL of 0.01 M NaCl and 50 mL of M `AgNO_(3)` solution. The solubility product of AgCl is `1.5xx10^(-10)`.

Options

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

C

Explanation

`NaCl+AgNO_(3)toAgCl+NaNO_(3)K_(sp)=1.5xx10^(-10)`
`[Ag^(+)]=(1)/(2)xx10^(-2)M=0.5xx10^(-2)M`
`[Cl^(-)]=(1)/(2)xx10^(-2)M=0.5xx10^(-2)M`
`K_(ip)=[Ag^(+)][Cl^(-)]=(0.5xx10^(-2))xx(0.5xx10^(-2))=2.5xx10^(-5)K_(ip)gtK_(sp)`
When `K_(ip)gtK_(sp)`, it results in precipitation.