# What is the molarity of the NaOH if 41.67 ml of this solution is needed to neutralize 0.2625 grams of pure 3H_2C_2O_4 + 2H_2O#?

Nov 29, 2016

$\textsf{0.3975 \textcolor{w h i t e}{x} \text{mol/l}}$

#### Explanation:

$\textsf{{H}_{2} {C}_{2} {O}_{4} + 2 N a O H \rightarrow N {a}_{2} {C}_{2} {O}_{4} + 2 {H}_{2} O}$

$\therefore$2 mol $\textsf{N a O H \equiv}$ 1 mol $\textsf{{H}_{2} {C}_{2} {O}_{4}}$

$\textsf{{n}_{{H}_{2} {C}_{2} {O}_{4}} = \frac{m}{M} _ r = \frac{0.2625}{90.035} = 0.0029155}$

$\therefore$$\textsf{{n}_{N a O H} = 0.0029155 \times 2 = 0.005831}$

$\textsf{c = \frac{n}{v} = \frac{0.005831}{0.01467} = 0.3975 \textcolor{w h i t e}{x} \text{mol/l}}$