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. 1.2 M
B. 1.8 M
C. 0.5 M
D. 0.4 M

Answer

A

Explanation

2 L of `3M AgNO_(3)` will contain 6 moles of ` AgNO_(3)`
`3L of 1 M BaCI_(2) "will contain 3 moles of" BaCI_(2)`
`2AgNO_(3) + BaCI_(2) rarr2 AgCI + Ba(NO)_(2)`
`"Thus,6 moles of" AgNO_(3)` will react 3 moles of `BaCI_(2)`, i.e. the two solutions will react completely to from 3 moles of `Ba(NO_(3))_2`
equiv 6 moles of `NO_(3)^(-)` ionsin `2+3 =5L`
Hence ,molarity of `NO_(3)^(-)=M=n/(V(L))=6/5=1.2`