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

a. Both `K_(2) CO_(3)` and `Na_(2) CO_(3)` are basic and hence both of them will react with `HCl`.
Note that for experiment, `20 mL` sample is taken from `100 mL` of the solution.
Let `x =` mass of `Na_(2) CO_(3)` in the solution
Mass of `K_(2) CO_(3)` in the solution `= (1.22 - x) g`
Now mass of `Na_(2) CO_(3)` in `20 mL` of the sample `= (x)/(5)`
Mass of `K_(2) CO_(3)` in the sample `= (1.22 - x)/(5)`
At end point: mEq of mixture = mEq of `HCl`
`[(x//5)/(106//2) + ((1.22 - x)//5)/(138//2)] xx 1000 = 0.1 xx 40`
`x = 0.53 g =` Mass of `Na_(2) CO_(3)`
and Weight of `K_(2) CO_(3) = 1.22 - 0.53 = 0.69 g`
b. When `20 mL` of sample is treated with `BaCl_(2)`,
`BaCO_(3)` is precipitated.
`BaCl_(2) + Na_(2) CO_(3) rarr BaCO_(3) + 2 NACl`
`BaCl_(2) + K_(2) CO_(3) rarr BaCO_(2) + 2 KCl`
Mass of `Na_(2) CO_(3)` in `20 mL = x//5 = 0.53//5 = 0.106 g`
Mass of `K_(2) CO_(3)` in `20 mL = 0.69//5 = 0.138 g`
`implies` Moles of `Na_(2)CO_(3) = 0.106//106 = 1.0 xx 10^(-3)`
Mole of `K_(2) CO_(3) = 0.138//138 = 1.0 xx 10^(-3)`
From stochiometry of above reactions, we have moles of `BaCO_(3)=` mole of `Na_(2) CO_(3) + "Moles of" K_(2) CO_(3))`
`= 1.0 xx 10^(-3) + 1.0 xx 10^(-3)`
`= 2.0 xx 10^(-3)`
`implies` Mass of `BaCO_(3) = (2.0 xx 10^(-3)) xx 197.4 = 0.395 g`