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

Explanation

Suppose weight of `Na_(2)CO_(3)` in the mixture = xg `therefore` Weight of `K_(2)CO_(3)` in the mixture `=(1.2-x)g`
`"Eq. wt of "Na_(2)CO_(3)=(46+12+148)/(2)=53,`
`"Eq. wt. of "K_(2)CO_(3)=(78+12+48)/(2)=69`
`"No. of g eq. of "N_(2)CO_(3) and K_(2)CO_(3)" in the mixture "=(x)/(53)+(1.2-x)/(69)`
`"40 cc of 0.1 N HCl contain g eq. of HCl "=(0.1)/(1000)xx40=4xx10^(-3)`
`"Thus, 20 cc of the mixture sol neutralize HCl "=4xx10^(-3)"g eq."`
`therefore" 100 cc of the mixture sol. will neutralise HCl"=(4xx10^(-3))/(20)xx100=2xx10^(-2)"g eq=0.02 g eq."`
As substance react in equivalent amounts ,`(x)/(53)+(1.20-x)/(69)=0.02`
or `69x+63.6-53x=0.02xx53xx69=73.14 or 16x=9.54 or x=0.596g`
Thus, `Na_(2)CO_(3)=0.596g and K_(2)CO_(3)=1.2-0.596=0.604 g`