Solubility product of `AgCl` is `2.8xx10^(-10)` at `25^(@)C`. Calculate solubility of the salt in `0.1 M AgNO_(3)` solution

Options

A. `2.8 xx 10^(-9)" mole/litre"`
B. `2.8 xx 10^(-10)" mole/litre"`
C. `3.2 xx 10^(-9)" mole/litre"`
D. `3.2 xx 10^(-12)" mole/litre"`

Answer

A

Explanation

In 0.1 M `AgNO_(3)`
`AgNO_(3) Leftrightarrow Ag^(+) +NO_(3)^(-)`
`AgCl Leftrightarrow Ag^(+) +Cl^(-)`
`K_(SP)=[Ag^(+)] [Cl^(-)]`
Now `[Ag^+]` can be taken as `[AgNO^3]" while "[Cl^-]` is the solubility of AgCl
`therefore Cl =(K_(sp))/([Ag^(+)])= (2.8 xx 10^(-10))/(0.1)`
Solubility of AgCl `=2.8 xx 10^(-9)" mole/litre"`