5 mL of 8 N HNO_(3), 4.8 mL of 5 N HCl, and a certain volume of 17 m H_(2) SO_(4) are mixed together and made upto 2 L. 30 mL of the acid mixture exactly neutralises 42.9 mL of Na_(2 CO_(3) solution containing 0.1 g of Na_(2) CO_(3). 10 H_(2) O in 10 mL of water. Calculate: a. The volume of H_(2) SO_(4) added to the mixture. b. The amount (in g) of the sulphate ions in the solution.

5 mL of 8 N HNO_(3), 4.8 mL of 5 N HCl, and a certain volume of 17 m H

1.	5 mL of 8 N HNO_(3), 4.8 mL of 5 N HCl, and a certain volume of 17 m H_(2) SO_(4) are mixed together and made upto 2 L. 30 mL of the acid mixture exactly neutralises 42.9 mL of Na_(2 CO_(3) solution containing 0.1 g of Na_(2) CO_(3). 10 H_(2) O in 10 mL of water. Calculate: a. The volume of H_(2) SO_(4) added to the mixture. b. The amount (in g) of the sulphate ions in the solution.
Answer» Solution :mEq of acid mixture `= mEq of HNO_(3) + mEq of HCl + mEq of H_(2) SO_(4)` LET `N` be the normality of the acid mixture and `V mL` be the volume of `H_(2) SO_(4)` added. `N xx 200= 8 xx 5 xx 4.8 + 2 (17) xx V` Now find `N` of carbonate as follows: `implies N = ("Strength")/(Ew)` Strength `= 0.1//10 mL -= 10 g L^(-1)` `Ew = Mw//2 = 286//2 = 143` (`Mw = 106 + 180`, adding the mass of `10 H_(2) O`) `N = (10)/(143)` Now mEq of acid mixture = mEq of `Na_(2)CO_(3)` solution `N xx 30 = 42.9 xx (10)/(143)` `implies N = 0.1 =` Normality of acid mixture Substituting in equation (i) we get: `0.1 xx 2000 = (40 + 24 + 34) V_(mL)` For grams of sulphate ions: mEq of `H_(2) SO_(4) = 2 xx 17 xx V = 136``(VmL = 4)` Now, mEq of `SO_(4)^(2-) = mEq of H_(2) SO_(4)` `("Weight")/(E_(SO_(4)^(2-))) xx 1000 = 136` `("Weight")/(96//2) xx 1000 = 136` Weight = Grams of `SO_(4)^(2-)` ions `= 6.53 g`