`1.20` gm sample of `NaCO_(3) " and " K_(2)CO_(3)` was dissolved in water to form 100 ml of a Solution : 20 ml of this solution required 40 ml of `0.1` N HCl for complete neutralization .
Calculate the weight of `Na_(2)CO_(3)` in the mixture. If another 20 ml of this solution is treated with excess of `BaCl_(2)` what will be the weight of the precipitate ?

Explanation

Let, weight of `Na_(2)CO_(3) = x gm`
Weight of `K_(2)CO_(3) = y ` gm
` :. X + y = 1.20` gm
For neutralization reaction of 100 ml
Meq. of `Na_(2)CO_(3) + " Meq. of " K _(2)CO_(3) = ` Meq. Of HCl
`rArr x/(106) xx 2 xx 1000 + y/(138) xx 2 xx 1000 = (40 xx 0.1 xx 100)/20 `
` :. 69x + 53y = 73 .14`
From Eqs. (1) and (2) , we get
`x = 0.5962 ` gm
` y = 0.604` gm
Solution of `Na_(2)CO_(3) " and " K_(2)CO_(3) ` gives ppt. of `BaCO_(3)` with `BaCl_(2)` (Meq. of `Na_(2)CO_(3) + ` Meq. of `K_(2)CO_(3))` in 20 ml = Meq. of `BaCO_(3)`
`rArr ` Meq. of HCl for 20 ml mixture = Meq. of `BaCO_(3)`
`rArr` Meq. of `BaCO_(3) = 40 xx 0.1 = 4`
` (197)/2 = W_(BaCO_(3))/E_(BaCO_(3)) xx 1000 = 40 xx 0.1 = 4`
`W_(BaCO_(3))/(197) xx 2 xx 1000 = 4`
` :. W_(BaCO_(3)) = 0.394` gm