# Question #b6f4c

Sep 12, 2017

Approx. $10 \cdot m L$

#### Explanation:

With all these problems, the priority is to write a stoichiometrically balanced equation:

${H}_{2} S {O}_{4} \left(a q\right) + 2 N a O H \left(a q\right) \rightarrow N {a}_{2} S {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

And thus ONE equiv of sulfuric acid reacts with TWO equiv of sodium hydroxide.

$\text{Moles of sodium hydroxide} = 60.2 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.427 \cdot m o l \cdot {L}^{-} 1 = 0.0257 \cdot m o l$.

And thus $\frac{0.0257 \cdot m o l}{2}$ ${H}_{2} S {O}_{4}$ are required.....

And so we take the quotient....

$\frac{\frac{0.0257 \cdot m o l}{2}}{1.28 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1 = 10.0 \cdot m L$

Note that normally, when we do a titration we aim to use a volume of $15 - 20 \cdot m L$ in order to minimize errors. Of course, you gots no choice when solving a problem.....