Question

Oxalic acid H2C2O4 is often used as a primary standard. Oxalic acid is a diprotic weak acid. You determine 1.2335 g of oxalic acid neutralizes 58.04 mL of an unknown concentration of NaOH solution.

1) Write the balance equation between oxalic acid and NaOH.

2) Calculate the molar connection of the NaOH solution.

Answer 1

1)

Balanced chemical equation is:

H2C2O4 + 2 NaOH ---> Na2C2O4 + 2 H2O

2)

Molar mass of H2C2O4,

MM = 2*MM(H) + 2*MM(C) + 4*MM(O)

= 2*1.008 + 2*12.01 + 4*16.0

= 90.036 g/mol

mass(H2C2O4)= 1.2335 g

use:

number of mol of H2C2O4,

n = mass of H2C2O4/molar mass of H2C2O4

=(1.234 g)/(90.04 g/mol)

= 1.37*10^-2 mol

According to balanced equation

mol of NaOH reacted = (2/1)* moles of H2C2O4

= (2/1)*1.37*10^-2

= 2.74*10^-2 mol

This is number of moles of NaOH

volume , V = 58.04 mL

= 5.804*10^-2 L

use:

Molarity,

M = number of mol / volume in L

= 2.74*10^-2/5.804*10^-2

= 0.4721 M

Answer: 0.472 M