# How many "mL" of "H"_2"C"_2"O"_4 can be neutralized by "23.65 mL" of "1.115 M" "NaOH"?

##### 1 Answer
Dec 5, 2014

The volume of oxalic acid required is $6.75 m L$.

First, start with the balanced chemical equation for this particular neutralization reaction

${H}_{2} {C}_{2} {O}_{4} \left(a q\right) + 2 N a O H \left(a q\right) \to N {a}_{2} {C}_{2} {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

From the chemical reaction we can see that $1$ mole of ${H}_{2} {C}_{2} {O}_{4}$ needs $2$ moles of $N a O H$ in order to produce $1$ mole of $N a {C}_{2} {O}_{4}$ and $2$ moles of ${H}_{2} O$.

We can determine the number of moles of $N a O H$ by using

${n}_{N a O H} = C \cdot V = 1.115 \frac{m o l e s}{L} \cdot 0.02365 L = 0.026 m o l e s$

This means that we need

${n}_{{H}_{2} {C}_{2} {O}_{4}} = \frac{{n}_{N a O H}}{2} = \frac{0.026}{2} = 0.013 m o l e s$ of oxalic acid.

Therefore, the volume of oxalic acid can be calculated from

${C}_{{H}_{2} {C}_{2} {O}_{4}} = {n}_{{H}_{2} {C}_{2} {O}_{4}} / {V}_{{H}_{2} {C}_{2} {O}_{4}}$

${V}_{{H}_{2} {C}_{2} {O}_{4}} = \frac{0.013 m o l e s}{1.978 \frac{m o l e s}{L}} = 6.57 m L$