`0.50 g` of a mixture of `K_(2)CO_(3)` and `Li_(2)CO_(3)` required `30 mL` of `0.25 N HCl` solution for neutralization. What is `%` composition of mixure?

Explanation

Weight of `K_(2)CO_(3) = a g`, weight of `Li_(2)CO_(3) = b g`
`:. a+b = 0.5` …(1)
Also for reaction:
Meq. of `K_(2)CO_(3) + "Meq. of" Li_(2)CO_(3) = "Meq. of" HCl`
`(a xx 1000)/(138//2) + (b xx 1000)/(74//2) = 30 xx 0.25`
`:. 74 a + 138b = 38.295` ...(2)
By eqs. (1) and (2), `a = 0.48 g, b = 0.02 g`
`:. %` of `K_(2)CO_(3) = (0.48)/(0.5) xx 100 = 96%`
`%` of `Li_(2)CO_(3) = (0.02)/(0.5) xx 100 = 4%`