Answer
$V(KOH)=38.8ml$
Work Step by Step
$c(KOH)=0.150\frac{mol}{dm^{3}}$
$V(H_{2}SO_{4})=29.1ml=0.0291dm^{3}$
$c(H_{2}SO_{4})=0.100\frac{mol}{dm^{3}}$---$V(KOH)=?$
The neutralization reaction of $H_{2}SO_{4}$ by $KOH$:
$2KOH+H_{2}SO_{4} \rightarrow K_{2}SO_{4}+2H_{2}O$
We can observe that the amount of potassium hydroxide required for the complete neutralization is twice the amount of sulfuric acid (according to the reaction's coefficients).
$n(KOH)=2n(H_{2}SO_{4})=2c(H_{2}SO_{4})\times V(H_{2}SO_{4})=2\times 0.100\frac{mol}{dm^{3}}\times 0.0291dm^{3}=5.82\times 10^{-3}mol$
Now, when we have calculated the amount of potassium hydroxide required for the complete neutralization, we can obtain the required volume of the solution by using its concentration:
$V(KOH)=\frac{n(KOH)}{c(KOH)}=\frac{5.82\times 10^{-3}mol}{0.150\frac{mol}{dm^{3}}}=0.0388 dm^{3}=38.8ml$
