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.

1.2 g mixture of Na_(2)CO_(3) and K_(2)CO_(3) was dissolved in water t

1.	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.
Answer» Solution :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`