500 mL fo 2 M HCl, 100 mL of 2 M H_(2) SO_(4), and one gram equivalent of a monoacidic alkali are mixed together. 30mL of this solution requried 20 mL of 143 g Na_(2) CO_(3). xH_(2)O in one litre solution. Calculate the water of crystallisation of Na_(2) CO_(3). xH_(2) O

500 mL fo 2 M HCl, 100 mL of 2 M H_(2) SO_(4), and one gram equivalent

1.	500 mL fo 2 M HCl, 100 mL of 2 M H_(2) SO_(4), and one gram equivalent of a monoacidic alkali are mixed together. 30mL of this solution requried 20 mL of 143 g Na_(2) CO_(3). xH_(2)O in one litre solution. Calculate the water of crystallisation of Na_(2) CO_(3). xH_(2) O
Answer» Solution :i. TOTAL mEq of acid `= 500 xx 2 + 100 xx 2 xx 2` (`n` factor) `= 1000 + 400 = 1400` Total mEq of monoacid alkali = 1000 Acid left after neutralisation with alkali `= 1400 - 1000 = 400 mEq` `N` fo acid left `= (400 mEq)/((500 + 100)mL) = (400)/(600) = 0.6 N` mEq of acid `-= mEq "of"Na_(2)CO_(3)` `30 mL xx 0.66 N = 20 mL xx N Na_(2)CO_(3)` Strength of `Na_(2)CO_(3) x H_(2)O = 143 g L^(-1)` `M_(Na_(2)CO_(3).xH_(2)O) = ("Strenght")/(Mw) = ((143)/(106 + 18x))` `M_(Na_(2)CO_(3).xH_(2)O) = ((2 xx 143)/(106 + 18 x))` (`n` factor = 2) Substitution the value of `N` of `Na_(2)CO_(3).xH_(2)O` in EQUATION (i), we get `30 xx 0.66 x = 20 xx ((2 xx 143)/(106 + 18 x))` solve for `x` `x = 10.16 ~~ 10` So WATER of crystallisation of `Na_(2)CO_(3).xH_(2) O = 10` formula : `Na_(2) CO_(3). 10 H_(2) O`