# A 40.16*mL volume of sodium hydroxide solution with concentration of 0.1693*g*mol^-1 reached an endpoint with a quantity of oxalic acid. If the oxalic acid was initially dissolved in a 30.00*mL what was the concentration of oxalic acid?

Nov 23, 2015

Concentration of oxalic acid $=$ $0.1133$ $m o l \cdot {L}^{-} 1$.

#### Explanation:

$H O \left(O =\right) C - C \left(= O\right) O H + 2 N a O H \rightarrow {\left\{C \left(= O\right) O\right\}}_{2}^{2 -} N {a}_{2}^{+} + 2 {H}_{2} O$

Moles of $N a O H$ for equivalence $=$ $40.16 \times {10}^{-} 3 L \times 0.1693$ $m o l \cdot {L}^{-} 1 =$ $6.80 \times {10}^{-} 3$ $m o l$.

Thus there were $3.40 \times {10}^{-} 3$ $m o l$ oxalic acid (because each equiv acid contributes 2 equiv ${H}^{+}$).

Since this quantity was dissolved in a $30.00$ $m L$ volume, concentration was $\frac{3.40 \times {10}^{-} 3 \cdot m o l}{30.00 \times {10}^{- 3} L}$ $=$ $0.1133$ $m o l \cdot {L}^{-} 1$ in oxalic acid.