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

Options

A. `2.5xx10^(-9)`
B. `2.5xx10^(-7)`
C. `2.5xx10^(-8)`
D. `2.5xx10^(-8)`

Answer

1

Explanation

`[Ag^(+)]=(10^(-3))/(50+50)xx1000=0.01`
`[Cl^(-)]=(0.1xx50)/(50+50)=0.05`
`Q=[Ag^(+)][Cl^(-)]=5xx10^(-4)`
`:' Q gtK_(sp)`
`{:(,Ag^(+)(aq.),+,Cl^(-)(aq.),rarr,AgCl(s)),("at given condition",0.01,,0.05,,),("after precipitation",x,~=,0.04,,):}`
`K_(sp)=[Ag^(+)][Cl^(-)],10^(-10)=x xx0.04`
`x=[Ag^(+)]=2.5xx10^(-9)`