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`)

Answer

`Na_(2)CO_(3)=0.592g, K_(2)CO_(3)=0.6038g`
`BaCO_(3)=0.394g`