A mixture containing only `Na_(2) CO_(3)` and `K_(2) CO_(3)` and weighing `1.22 g` was dissolved in water to form `100 mL` of solution: `20 mL` of this solution required `40 mL` of `0.1 N HCl` for neutralisation.
a. Calculate the weight of `K_(2) CO_(3)` in the mixture.
b. If another `20 mL` of the same solution is treated with excess of `BaCl_(2)`, what will be the weight of precipitate thus obtained? (Molarcular of `Na_(2)CO_(3) = 106`, `K_(2) CO_(3) = 138, BaCO_(3) = 197.4`)

Explanation

Weight of `Na_(2)CO_(3) = a g`
Weight of `K_(2)CO_(3) = b g`
`:. a+b = 1.20` …(1)
For neutralization reaction of `100 mL` solution
Meq. Of `Na_(2)CO_(3) + "Meq. of" K_(2)CO_(3) = "Meq.of" HCl`
`(a)/(106//2) xx 1000 + (b)/(138//2) xx 1000 = (40 xx 0.1 xx 100)/(20)`
`:. 69a+53b = 73.14` ...(2)
By eqs. (1) and (2), `a = 0.5962 g, b = 0.6038 g`
Further, solution of `Na_(2)CO_(3) + K_(2)CO_(3)` gives ppt. of `BaCO_(3)` with `BaCl_(2)`.
`"Meq. of" BaCO_(3) = "Meq. of" Na_(2)CO_(3)+"Meq. of"K_(2)CO_(3)`
(in `20 mL`)
`= "Meq. of HCl "for" 20mL "mixture" = 40 xx 0.1 = 4`
`:. (w)/(197//2) xx 1000 = 4`
`:.` Wt. of `BaCO_(3) = 0.394 g`