Answer
$c(NaOH) = 0.0310 M$
Work Step by Step
Considering stoichiometric coefficients, we can conclude that the amount of sodium - hydroxide required is equal to the amount of nitric acid.
$n(NaOH) = n(HNO_{3}) = c(HNO_{3})\times V(HNO_{3}) = 0.0342\frac{mol}{dm^3}\times 0.0375 dm^3 = 0.0012825mol$
Hence, the molarity of $NaOH$ solution is equal to:
$c(NaOH) = \frac{n(NaOH)}{V(NaOH)} = \frac{0.0012825mol}{0.0414dm^3} = 0.0310 M$
