In what ratio should you mix `0.2 M NaNO_(3)` and `0.1 M Ca (NO_(3))_(2)` solution so that in resulting solution the concentration of -ve ion is 50 % greater than the concentration of + ve ions?

Answer

2

Explanation

Let `v_(1)` be the volume of `Ca(NO_(3))_(2)` solution
and `v_(2)` be the volume of `NaNO_(3)` solution
concentration of anion `(NO_(3)^(-))` in the mixture `= ((2 xx M_(1)V_(1)) + M_(2)V_(2))/(V_(1) + V_(2))`
concentration of cation `(Ca^(2+), Na^(+))` in the mixture `= (M_(1)V_(1) + M_(2)V_(2))/(V_(1) + V_(2))`
conc of anion = conc of cation `+ (50)/(100)` (conc of cation)
`= (3)/(2) xx "conc of cation " :. " " (2M_(1)V_(1) + M_(2)V_(2))/(V_(1)+V_(2)) = (3)/(2)(M_(1)V_(1) + M_(2)V_(2))/(V_(1) + V_(2))`
`rArr 2(0.1)V_(1) + 0.2 V_(2) = (3)/(2) (0.1 V_(1) + 0.2 V_(2)) rArr 0.4 V_(1) + 0.4 V_(2) = 0.3 V_(1) + 0.6 V_(2)`
`rArr 0.1 V_(1) 0.2 V_(2) rArr (V_(1))/(V_(2)) = 2`