Question

A chemistry student needs to standardize a fresh solution of sodium hydroxide. She carefully weights out 305.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 69.4mL 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

Given mass of oxalic acid is 305 mg = 305x10^-3 g

Molar mass of oxalic acid , H2C2O4 = (2x1) + (2x12)+(4x16) = 90 g/mol

So number of moles of oxalic acid , n = mass/ molar mass

                                                       = (305x10^-3 g )/ 90 (g/mol)

                                                        = 3.39x10^-3 mol

Given volume of oxalic acid solution , V = 250.0 mL = 0.250 L

Therefore Molarity of thre oxalic acid solution , M = Number of moles / Volume of solution in L

                                                                       = (3.39x10^-3 mol) / 0.250 L

                                                                       = 0.0135 M

The balanced reaction between oxalic acid & NaOH is

H₂C₂O₄ + 2NaOH Na₂C₂O₄ + 2H₂O

According to the above balanced equation ,

1 mole of oxalic acid reacts with 2 moles of NaOH

3.39x10^-3 mol of oxalic acid reacts with 2x3.39x10^-3 mol= 6.78x10^-3 moles of NaOH

So Molarity of the unknown NaOH solution , M = number of moles of NaOH reacted / Volume of NaOH solution in L

                                                                     = 6.78x10^-3 mol / (69.4 /1000) L

                                                                    = 0.098 M

Therefore the molarity of unknown NaOH solution = 0.098 M