# Question 21b47

##### 1 Answer
Mar 19, 2017

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).

Unless you provide actual numbers, this is as far as we can go.