`50mL` of a solution containing `10^(-3)` mole of `Ag^(+)` is mixed with `50 mL` of a `0.1M HCl` solution. How much `Ag^(+)` remains in solution ? `(K_(sp)` of `AgCl=1.0xx10^(-10))`

Explanation

Moles of `Cl^-` ions present in 50 mL of 0.10 M HCl =`0.10/1000xx50`=0.005 mol
Let x mol of `Ag^+` ion remains in solution.
Moles of `Ag^+` ion precipitated =0.001-x `approx` 0.001 mol ( x is very small)
Moles of `Cl^-` ion precipitated =0.001 mol
Moles of `Cl^-` left =0.005-0.001 =0.004 mol
Total volume of solution =50+50=100 mL
`[Cl^-]=0.004 /100xx1000`=0.040 M
`K_(sp)=[Ag^+][Cl^-]=[Ag^+]xx0.040=1xx10^(-10)`
`therefore [Ag^+]=(1xx10^(-10))/0.04=2.5xx10^(-9)`
Moles of `Ag^+` ions remaining in 100 mL
`=(2.5xx10^(-9))/1000xx100`
`=2.5xx10^(-10)` mol