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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» <html><body><p></p>Solution :mEq of acid mixture <br/> `= mEq of HNO_(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) + mEq of HCl + mEq of H_(2) SO_(4)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> `N` be the normality of the acid mixture and `<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> mL` be the volume of `H_(2) SO_(4)` added. <br/> `N xx 200= 8 xx 5 xx 4.8 + 2 (17) xx V` <br/> Now find `N` of carbonate as follows: <br/> `implies N = ("Strength")/(Ew)` <br/> Strength `= 0.1//10 mL -= 10 g L^(-1)` <br/> `Ew = Mw//2 = 286//2 = 143` <br/> (`Mw = <a href="https://interviewquestions.tuteehub.com/tag/106-266627" style="font-weight:bold;" target="_blank" title="Click to know more about 106">106</a> + 180`, adding the mass of `10 H_(2) O`) <br/> `N = (10)/(143)` <br/> Now mEq of acid mixture = mEq of `Na_(2)CO_(3)` solution `N xx 30 = 42.9 xx (10)/(143)` <br/> `implies N = 0.1 =` Normality of acid mixture <br/> Substituting in equation (i) we get: <br/> `0.1 xx 2000 = (<a href="https://interviewquestions.tuteehub.com/tag/40-314386" style="font-weight:bold;" target="_blank" title="Click to know more about 40">40</a> + 24 + 34) V_(mL)` <br/> For grams of sulphate ions: <br/> mEq of `H_(2) SO_(4) = 2 xx 17 xx V = 136``(VmL = 4)` <br/> Now, mEq of `SO_(4)^(2-) = mEq of H_(2) SO_(4)` <br/> `("Weight")/(E_(SO_(4)^(2-))) xx 1000 = 136` <br/> `("Weight")/(96//2) xx 1000 = 136` <br/> Weight = Grams of `SO_(4)^(2-)` ions `= 6.53 g`</body></html>


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