# 3.6 g of oxalic acid (H_2C_2O_4 * 2H_2O) are dissolved and the volume is made upto 150 ml. Calculate the normality of the solution for redox titration?

Jul 12, 2015

Normality: 0.38 N.

#### Explanation:

The purpose of using oxalic acid dihydrate, ${H}_{2} {C}_{2} {O}_{4} \cdot 2 {H}_{2} O$, in a redox titration is that the oxalate ion, ${C}_{2} {O}_{4}^{2 -}$, will act as a reducing agent and reduce, for example, potassium permanganate, $K M n {O}_{4}$, to the $M {n}^{2 +}$ ion.

In order to determine the normality of the oxalic acid solution, you need to first figure out how many electrons will the reducing agent lose.

These electrons will help you determine the equivalents needed to calculate the solution's normality.

Now, I won't show you exactly how to balance redox reactions because I don't want the answer to become too long. The balanced chemical equation for the redox titration of potassium permanganate with oxalic acid looks like this

$6 {H}_{\left(a q\right)}^{+} + 5 {C}_{2} {O}_{4 \left(a q\right)}^{2 -} + 2 M n {O}_{4 \left(a q\right)}^{-} \to 2 M {n}_{\left(a q\right)}^{2 +} + 10 C {O}_{2 \left(g\right)} + 8 {H}_{2} {O}_{\left(l\right)}$

I will show you how to balance the oxidation half-reaction, in which the oxalate ion, ${C}_{2} {O}_{4}^{2 -}$, is oxidized to carbon dioxide.

This half-reaction will tell you how many electrons are being lost by the reducing agent.

stackrel(color(blue)(+3))(C_2) O_4""^(2-) -> stackrel(color(blue)(+4))(C)O_2

The carbon atoms are going from an oxidation state of +3 on the reactants' side, to an oxidation state of +4 on the products' side.

However, keep in mind that you have 2 carbon atoms on the reactants' side, so multiply the carbon dioxide by 2 to get

stackrel(color(blue)(+3))(C_2) O_4""^(2-) -> 2stackrel(color(blue)(+4))(C)O_2

This means that a total of 2 electrons are being lost, one from each of the two carbon atoms.

stackrel(color(blue)(+3))(C_2) O_4""^(2-) -> 2stackrel(color(blue)(+4))(C)O_2 + 2e^(-)

This tells you that 1 mole of oxalic acid produces 2 moles of electrons for the redox reaction. These are your equivalents. You thus have

$\text{1 mole "H_2C_2O_4 * 2H_2O = "2 moles equivalents}$

Now, in order to determine how many moles of oxalic acid you have, you need to use the percent composition of the dihydrate form.

Oxalic acid dihydrate contains

(90.03cancel("g/mol"))/(126.07cancel("g/mol")) * 100 = "71.4% oxalic acid"

by mass. This means that your 3.6-g sample of dihydrate actually contains

3.6cancel("g dihydrate") * "71.4 g oxalic acid"/(100cancel("g dihydrate")) = "2.57 g" ${C}_{2} {H}_{2} {O}_{4}$

The number of moles of oxalic acid will be

2.57cancel("g") * "1 mole"/(90.03 cancel("g")) = "0.0285 moles"

This means that you get

0.0285cancel("moles") * "2 moles of eq."/(1cancel("mole")) = "0.057 moles eq."

The solution's normality will thus be

$N = \frac{\text{no. of moles eq.}}{V}$

N = "0.057 moles eq."/(150 * 10^(-3)"L") = color(green)("0.38 N")