Calculate the molarity of the student's sodium hydroxide solution. Be sure your answer has the correct number of significant digits.

Question

A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weighs out 68.mg

 of oxalic acid H2C2O4

, a diprotic acid that can be purchased inexpensively in high purity, and dissolves it in 250.mL

 of distilled water. The student then titrates the oxalic acid solution with her sodium hydroxide solution. When the titration reaches the equivalence point, the student finds she has used 64.7mL

 of sodium hydroxide solution.

Calculate the molarity of the student's sodium hydroxide solution. Be sure your answer has the correct number of significant digits.

Answer 1

At equivalence point moles of H⁺ = moles of OH^- n_H = n_OH

2n_OA = n_NaOH (OA contributes 2 moles of H⁺ per mole of OA

2m_OA/M_OA = C_NaOHV_NaOH where m is mass, M is molar mass, C is concentration and V is volume

C_NaOH = 2m_OA/(M_OAV_NaOH)

Now plug in, paying attention to units. The volume of NaOH needs to be in liters so that C will be moles/liter. You can work out the molar mass of oxalic acid (BTW, OA is used because it is a hydrate and doesn't take in water with time. For a real lab, you would use H₂C₂O₄•2H₂O, not the anhydrous version). Notice that the volume of the acid solution is irrelevant to the problem when you know the mass of solute.

Please consider a tutor. Take care.

Answer 2

I do believe that you asked the same question (with different amounts of material), just the other day. I answered it then, and will do so again, but you really should be able to do these yourself after reviewing my previous answer. Unless you didn't understand any or part of that answer. If so, you should state so.

H₂C₂O₄ + 2NaOH ==> Na₂C₂O₄ + 2H₂O ... balanced equation for the reaction taking place

Molar mass oxalic acid = 90.03 g / mol

moles H₂C₂O₄ present = 68 mg x 1 g /1000 mg x 1 mol / 90.03 g = 7.55x10^-4 moles

moles NaOH needed = 7.55x10^-4 mol H₂C₂O₄ x 2 mol NaOH / mol H₂C₂O₄ = 1.51x10^-3 mol NaOH

Concentration of NaOH = 1.51x10^-3 mol / 64.7 ml x 1000 ml / L = 0.0233 mols / L = 0.0233 M