What is the molarity of a sodium hydroxide solution if 40.00 mL of NaOH is required to neutralize 0.900 g of oxalic acid,H₂C₂O₄, (90.04 g/mol)? H₂C₂O₄(aq)+ 2 NaOH(aq)→Na₂C₂O₄(aq)+ 2 H₂O(l)

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First, calculate the moles of oxalic acid: $0.900 \, \text{g} \times \frac{1 \, \text{mol}}{90.04 \, \text{g/mol}} = 0.00999 \, \text{mol}$. Since the reaction requires 2 moles of NaOH for every mole of oxalic acid, the moles of NaOH needed are $0.00999 \, \text{mol} \times 2 = 0.01998 \, \text{mol}$. The molarity of NaOH is then calculated by dividing the moles of NaOH by the volume of the solution in liters: $0.01998 \, \text{mol} \div 0.04000 \, \text{L} = 0.4995 \, \text{M}$, which rounds to $0.500 \, \text{M}$.

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