Answer
$c(H_{2}SO_{4})=0.0945\frac{mol}{dm^{3}}$
Work Step by Step
$V(H_{2}SO_{4})=35.38ml=0.03538dm^{3}$
$m(Na_{2}CO_{3})=0.3545g$---$c(H_{2}SO_{4})=?$
The reaction between $Na_{2}CO_{3}$ and $H_{2}SO_{4}$:
$H_{2}SO_{4}+Na_{2}CO_{3}→Na_{2}SO_{4}+CO_{2}+H_{2}O$
We can observe that the amount of sulfuric acid required for the complete reaction with sodium carbonate is equal to the amount of sodium carbonate (according to the reaction's coefficients).
$n(H_{2}SO_{4})=n(Na_{2}CO_{3})=\frac{m(Na_{2}CO_{3})}{M(Na_{2}CO_{3})}=\frac{0.3545g}{106\frac{g}{mol}}=3.344×10^{−3}mol$
Now, when we have calculated the amount of sulfuric acid required for the complete reaction with sodium carbonate, we can obtain the required concentration of the solution by using its volume:
$c(H_{2}SO_{4})=\frac{n(H_{2}SO_{4})}{V(H_{2}SO_{4})}=\frac{3.344×10^{−3}mol}{0.03538dm^{3}}=0.0945\frac{mol}{dm^{3}}$
