Answer
$V(NaOH) = 22.85ml$
Work Step by Step
The following neutralization reaction takes place:
$2NaOH + H_{2}SO_{4} \rightarrow Na_{2}SO_{4} + 2H_{2}O$
The amount of sodium hydroxide required is twice the amount of sulfuric acid. Hence,
$n(NaOH) = 2n(H_{2}SO_{4}) = 2c(H_{2}SO_{4})V(H_{2}SO_{4}) = 2\times 0.234\frac{mol}{dm^3}\times 0.025 dm^3 = 0.0117mol$
Given the concentration of $NaOH$ solution, we can easily obtain its volume:
$V(NaOH) = \frac{n(NaOH)}{c(NaOH)} = \frac{0.0117mol}{0.512\frac{mol}{dm^3}}=0.02285dm^3=22.85ml$
