Calculate the molarity of each ion in solution after 2.0 litre of 3.0 M `AgNO_(3)` is mixed with 3.0 litre of 1.0 M `BaCl_(2)`.

Options

A. `[Ba^(2+)]=0.6 M, [NO_(3)^(-)]=1.2 M`
B. `[Ag^(+)]=[Cl^(-)]=1.2 M`
C. `[Ba^(2+)]=1.0 M, [NO_(3)^(-)]=3.0 M`
D. None of these

Answer

A

Explanation

`2AgNO_(3)+BaCl_(2) to 2AgCl(s)+Ba(NO_(3))_(2)(aq)`
`[Ba^(2+)]=(1.0 xx 3)/(2+3)=0.6 M`
`[NO_(3)^(-)]=(2 xx 1 xx 3)/(2+3)=1.2 M`
Moles of `Ag^(+)` mixed = `3 xx 2`=6
Moles of `Cl^(-)` mixed =`2 xx 1 xx 3`=6
Almost all `Ag^(+)` and `Cl^(-)` will be precipitated out to give 6 mol AgCl.