$5 ext{ mL}$ of $8 ext{ N HNO}_3$, $4.8 ext{ mL}$ of $5 ext{ N HCl}$ and a certain volume $V

Question

$5 	ext{ mL}$ of $8 	ext{ N HNO}_3$, $4.8 	ext{ mL}$ of $5 	ext{ N HCl}$ and a certain volume $V$ of $17 	ext{ M H}_2	ext{SO}_4$ are mixed together and made up to $2 	ext{ litres}$. $30 	ext{ mL}$ of this acid mixture exactly neutralizes $42.9 	ext{ mL}$ of sodium carbonate solution containing $1 	ext{ g}$ of $	ext{Na}_2	ext{CO}_3 \cdot 10	ext{H}_2	ext{O}$ dissolved in $100 	ext{ mL}$ of water. Calculate the number of millimoles of sulphate ions ($	ext{SO}_4^{2-}$) present in the final $2 	ext{ litre}$ solution. (Molar mass of $	ext{Na}_2	ext{CO}_3 \cdot 10	ext{H}_2	ext{O} = 286 	ext{ g/mol}$)

Professor PC Thomas Classes · Answer

Molar mass of $	ext{Na}_2	ext{CO}_3 \cdot 10	ext{H}_2	ext{O} = 286$. Moles in $100 	ext{ mL} = 1/286$. Molarity $= 10/286 	ext{ M}$. Normality $= 20/286 = 10/143 	ext{ N}$. 
Equivalents in $42.9 	ext{ mL} = 42.9 	imes (10/143) = 3 	ext{ meq}$. 
Thus, $30 	ext{ mL}$ of the acid mixture contains $3 	ext{ meq}$ of acid. 
The $2000 	ext{ mL}$ mixture contains $(3/30) 	imes 2000 = 200 	ext{ meq}$ of acid. 
Total meq $= 	ext{meq}(	ext{HNO}_3) + 	ext{meq}(	ext{HCl}) + 	ext{meq}(	ext{H}_2	ext{SO}_4) = 200$. 
$5 	imes 8 + 4.8 	imes 5 + 	ext{meq}(	ext{H}_2	ext{SO}_4) = 200 \implies 40 + 24 + 	ext{meq}(	ext{H}_2	ext{SO}_4) = 200 \implies 	ext{meq}(	ext{H}_2	ext{SO}_4) = 136 	ext{ meq}$. 
Since the n-factor of $	ext{H}_2	ext{SO}_4$ is 2, moles of $	ext{H}_2	ext{SO}_4 = 136 / 2 = 68 	ext{ mmol}$. 
Thus, the number of millimoles of sulphate ions is $68$.