5 mLof 8 N nitricacid , 4.8 mLof 5 HCl and a certain volume of 17M sulphuric acid are mixed togther and made up to 2 litres . 30 mL of this mixture exactly neutralises 42.9 mL of sodium carbonate solution containing 1 g of Na_(2)CO_(3) . 10 H_(2)O in 100 mL of water . Calculate the amount in grams of the sulphate ions on the solution .

5 mLof 8 N nitricacid , 4.8 mLof 5 HCl and a certain volume of 17M sul

1.	5 mLof 8 N nitricacid , 4.8 mLof 5 HCl and a certain volume of 17M sulphuric acid are mixed togther and made up to 2 litres . 30 mL of this mixture exactly neutralises 42.9 mL of sodium carbonate solution containing 1 g of Na_(2)CO_(3) . 10 H_(2)O in 100 mL of water . Calculate the amount in grams of the sulphate ions on the solution .
Answer» Solution :Let the volume of 17 M ( i.e ., 34 N) `H_(2)SO_(4)` solution be v mL . ` :. ` total m.e of the acid mixture = `8 xx 5 + 5 xx 4.8 + 34 v "" ..(Eqn.1)` = `(64 + 34v)` ` :.` normal of the mixture = `(m.e)/( " total volume (mL)") ""…(Eqn .1)` ` = (64 + 34 v)/(2000) ` m.e of 30 mL ofthis acidmixture` = (64+ 34v)/2000 xx 30` Now , NORMALITY of `Na_(2)CO_(3) . 10 H_(2)O ` solution = `(g//litre)/("eq.wt")` ` = 10/43 ` ` {{:( "grams/litre of "Na_(2)CO_(3).10H_(2)O=10),("and eq. wt"=(mol.wt)/2 = 286 /2 ):}}` ` :. ` m.eof 42.9 mL of `Na_(2)CO_(3).10H_(2)O ` solution = `10/143 xx 42.9` THUS , m.e of 30 mL of acid mixture = m.e of `42.9 " mL of " Na_(2)CO_(3) . 10 H_(2)O ` solution . ` :. ` m.eof N . (i.e ., 17 M) `H_(2)SO_(4) = 34 xx 68/17 "" ...(Eqn.1)` ` = 136 ` ` :.` equivalent of `H_(2)SO_(4) = 136/1000 = 0.136 "" ...(Eqn.3)` ` :. " equivalent of"SO_(4)^(2-) = 0.136 ""....(Eqn . 7II)` weight of `SO_(4)^(2-)= eq xx eq .wt of SO_(4)^(2-) "" ....(Eqn.4i)` ` = 0.136 xx 48 ` ` = 6.528 g ` `( " eq. wt of " SO_(4)^(2-)= (" ionix wt")/("valency")= 96/2 = 48 )`