1.20 gm sample of NaCO_(3) " and " K_(2)CO_(3) was dissolved in water to form 100 ml of a Solution : 20 ml of this solution required 40 ml of 0.1 N HCl for complete neutralization . Calculate the weight of Na_(2)CO_(3) in the mixture. If another 20 ml of this solution is treated with excess of BaCl_(2) what will be the weight of the precipitate ?

1.20 gm sample of NaCO_(3) " and " K_(2)CO_(3) was dissolved in water

1.	1.20 gm sample of NaCO_(3) " and " K_(2)CO_(3) was dissolved in water to form 100 ml of a Solution : 20 ml of this solution required 40 ml of 0.1 N HCl for complete neutralization . Calculate the weight of Na_(2)CO_(3) in the mixture. If another 20 ml of this solution is treated with excess of BaCl_(2) what will be the weight of the precipitate ?
Answer» Solution :Let, weight of `Na_(2)CO_(3) = xgm` Weight of `K_(2)CO_(3) = y ` gm ` :. X + y = 1.20` gm For neutralization reaction of 100 ml Meq. of `Na_(2)CO_(3) + " Meq. of" K _(2)CO_(3) = ` Meq. Of HCl `RARRX/(106) xx 2 xx 1000+ y/(138) xx 2 xx 1000 = (40 xx 0.1 xx 100)/20 ` ` :. 69x + 53Y = 73 .14` From Eqs. (1) and (2) , we get `x = 0.5962 ` gm ` y = 0.604` gm Solution of `Na_(2)CO_(3) " and " K_(2)CO_(3) ` gives ppt. of `BaCO_(3)` with `BaCl_(2)`(Meq. of `Na_(2)CO_(3) + ` Meq. of `K_(2)CO_(3))` in 20 ml = Meq. of `BaCO_(3)` `rArr ` Meq. of HCl for 20 ml mixture= Meq. of `BaCO_(3)` `rArr` Meq. of `BaCO_(3) = 40 xx 0.1 = 4` ` (197)/2= W_(BaCO_(3))/E_(BaCO_(3)) xx 1000 = 40 xx 0.1 = 4` `W_(BaCO_(3))/(197) xx 2 xx 1000= 4` ` :. W_(BaCO_(3)) = 0.394` gm