Answer
$V(H_{2}SO_{4})=12.75ml$
Work Step by Step
The following is the neutralization reaction of potassium hydroxide by sulfuric acid:
$2KOH + H_{2}SO_{4} \rightarrow K_{2}SO_{4} + 2H_{2}O$
By considering stoichiometric coefficients, we can observe that the amount of sulfuric acid required is two times less than the amount of potassium hydroxide. Hence,
$n(H_{2}SO_{4})=0.5\times n(KOH)=0.5\times c(KOH)\times V(KOH)=0.5\times 0.255\frac{mol}{dm^3}\times 0.0344dm^3=0.004386mol$
We can obtain the volume of $H_{2}SO_{4}$ solution knowing its concentration:
$V(H_{2}SO_{4})=\frac{n(H_{2}SO_{4})}{c(H_{2}SO_{4})}=\frac{0.004386mol}{0.344\frac{mol}{dm^3}}=0.01275dm^3=12.75ml$
