# A 10*mL volume of oxalic acid was titrated with a 34.40*mL volume of NaOH(aq) whose concentration was 0.250*mol*L^-1. What was the mass of oxalic acid present in the initial 10*mL volume?

Jul 31, 2017

$\left[{C}_{2} {O}_{4} {H}_{2}\right] \cong 0.5 \cdot m o l \cdot {L}^{-} 1$.

#### Explanation:

We need (i) a stoichiometric equation.....

$\text{HO(O=)CC(=O)OH + 2NaOH} \rightarrow {C}_{2} {O}_{4}^{2 -} N {a}_{2}^{+} + 2 {H}_{2} O$

And (ii) equivalent quantities of sodium hydroxide......

$\text{Moles of NaOH} = 34.40 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.250 \cdot m o l \cdot {L}^{-} 1 = 8.60 \times {10}^{-} 3 \cdot m o l$.

And HALF this molar quantity was equivalent to the oxalic acid solution contained in $10.00 \cdot m L$ solution.

$\text{Concentration of oxalic acid} = \frac{\frac{1}{2} \times 8.60 \times {10}^{-} 3 \cdot m o l}{10.00 \times {10}^{-} 3 \cdot {L}^{-} 1}$

$= 0.430 \cdot m o l \cdot {L}^{-} 1$

And thus in a $10 \cdot m L$ volume, there were $10 \times {10}^{-} 3 \cdot L \times 0.430 \cdot m o l \cdot {L}^{-} 1 \times 90.03 \cdot g \cdot m o {l}^{-} 1 =$

$\cong 0.4 \cdot g$ ""