Question

Consider the following two reactions involving oxalic acid.

2MnO4− + 5H2C2O4 + 6H+ → 2Mn2+ + 10CO2 + 8H2O

H2C2O4 + 2OH− → C2O42− + 2H2O

What volume of 0.0100 M KMnO₄ solution will titrate a solution prepared from 46.0 mg of H₂C₂O₄?

What volume of 0.0100 M NaOH solution will titrate a solution prepared from 46.0 mg of H₂C₂O₄?

Answer 1

1)

Balanced chemical equation is:

2MnO4− + 5H2C2O4 + 6H+ → 2Mn2+ + 10CO2 + 8H2O

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)= 0.046 g

use:

number of mol of H2C2O4,

n = mass of H2C2O4/molar mass of H2C2O4

=(4.6*10^-2 g)/(90.04 g/mol)

= 5.109*10^-4 mol

According to balanced equation

mol of MnO4- reacted = (2/5)* moles of H2C2O4

= (2/5)*5.109*10^-4

= 2.044*10^-4 mol

This is number of moles of MnO4-

use:

M = number of mol / volume in L

0.01 = 2.044*10^-4/ volume in L

volume = 0.0204 L

volume = 20.4 mL

Answer: 20.4 mL

2)

Balanced chemical equation is:

H2C2O4 + 2OH− → C2O42− + 2H2O

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)= 0.046 g

use:

number of mol of H2C2O4,

n = mass of H2C2O4/molar mass of H2C2O4

=(4.6*10^-2 g)/(90.04 g/mol)

= 5.109*10^-4 mol

According to balanced equation

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

= (2/1)*5.109*10^-4

= 1.022*10^-3 mol

This is number of moles of OH-

use:

M = number of mol / volume in L

0.01 = 1.022*10^-3/ volume in L

volume = 0.102 L

volume = 102 mL

Answer: 102 mL