# Question 21b47

Suppose $n$ $\text{mol}$s of $\text{NaOH}$ are contained in $\text{25 mL}$. Then,
$\text{n mols"/("25 mL") xx "10 mL} = 0.4 n$ $\text{mol}$s
That is, there is 40% of the $\text{mol}$s of $\text{NaOH}$ in 40%# of the volume, since $\text{mol}$s are an extensive quantity, just like $\text{grams}$.
Since $\text{NaOH}$ is a strong base, it reacts 1:1 with the first ${\text{H}}^{+}$ that oxalic acid dissociates into solution (we assume the second ${K}_{a}$ is negligible).