1.2 g mixture of `Na_(2)CO_(3) and K_(2)CO_(3)` was dissolved in water to form `100 cm^(3)` of the solution. `20 cm^(3)` of this solution required `40 cm6(3)` of 0.1 N HCl solution for neutralisation. Calculate the percentage composition of the mixture.

Explanation

Step I. Calculation of total no of gram equivalents in the mixture
Let the mass of `Na_(2)CO_(3)` in the mixture = x g
`:.` Mass of `K_(2)CO_(3)` in the mixture `= (1.2 - x)g`
Equivalent mass of `Na_(2)CO_(3)=("Molecular mass")/(2)=(106)/(2)=53 `
Equivalent mass of `K_(2)CO_(3)=("Molecular mass")/(2)=(138)/(3)=69`
No. of gram equivalents of `Na_(2)CO_(3)=(x)/(53)`
No. of gram equivalent of `K_(2)CO_(3)=((1.2 - x))/(69)`
Total no. of gram equivalent of `Na_(2)CO_(3)` and `K_(2)CO_(3) = (x)/(53) +((1.2 -x))/(69)`
Step II. Calculation of the no. of gram equivalents by titration
No. of gram equivalents in `40 cm^(3)` of 0.1 HCl `= (0.1)/(1000) xx 40 = 4 xx 10^(-3)`
`20 cm^(3)` of the mixture solution neutralise `HCl = 4 xx 10^(-3) g eq`.
`100 cm^(3)` of the mixture solution neutralise `HCl = (4 xx 10^(-3))/(20) xx 100 = 2 xx 10^(-2) g aq.`
Since acids and bases react in proportions of their gram equivalents
`:.` No. of gram equivalents of `Na_(2)CO_(3) and K_(2)CO_(3)=2 xx 10^(-2)`
Step III. Calculation of percentage composition of the mixture
Equating eqns. (i) and (ii), `(x)/(53)+((1.2-x))/(69)=2xx10^(-2)=0.02`
`69x+63.6-53x=0.02xx53xx69=73.14`
`16x=73.14-63.6=98.54`
`:. x =(9.54)/(16)=0.596g`
% of `Na_(2)CO_(3)` in the mixture `= (0.596)/(1.2) xx 100 = 49.67`
% of `K_(2)CO_(3)` in the mixture `= 100-49.67 = 50.33`.