Problem: Oxalic acid, H2C2O4•2H2O (molar mass = 126.07 g/mol) is often used as a primary standard for the standardization of a NaOH solution. If 0.147 g of oxalic acid dihydrate is neutralized by 23.64mL of a NaOH solution, what is the molar concentration of the NaOH solution? Oxalic acid is a diprotic acid. (What is the balanced equation?)

Step 1. Write the chemical equation for the reaction

H₂C₂O₄(aq) + 2 NaOH(aq) → 2 H₂O(l) + C₂O₄^2-(aq)

*Main group metals should have a +3 charge or higher to be considered acidic. Na⁺ has only a +1 charge and is a neutral ion so it will not contribute to the pH of the solution

Step 2. Calculate the moles of H₂C₂O₄

$\mathbf{moles}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{147}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{·}\mathbf{2}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{·}\mathbf{2}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}{\mathbf{126}\mathbf{.}\mathbf{07}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{·}\mathbf{2}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}\mathbf{·}\mathbf{2}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}$

moles H₂C₂O₄ = 1.166×10^‒3 mol

Step 3. Calculate the moles of NaOH using mole-to-mole comparison

$\mathbf{moles}\mathbf{ }\mathbf{NaOH}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{166}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}}\mathbf{\times }\frac{\mathbf{2}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{NaOH}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}}}$

moles NaOH = 2.332×10^‒3 mol

Step 4. Calculate the concentration of NaOH