A 15.00 mL sample of a solution containing oxalic acid, `H_2 C_2 O_4` was titrated with 0.02 M `KMnO_4`. The titration, required 18.30 mL of the `KMnO_4` solution. What was the molarity of the `H_2 C_2O_4` solution? In the reaction, oxalate ion `(C_2O_(4)^(2-))` is oxidized to `CO_2` and `MnO_(4)^(-)` is reduced to `Mn^(2+)`.

Options

A. `0.082`
B. `0.061`
C. `0.061`
D. `0.82`

Answer

B

Explanation

First we need a balanced equation : ` C_2 O_(4)^(2-) to CO_2 to CO_2 , MnO_(4)^(-) to Mn^(2+) to Mn^(2+)`
Moles of ` C_2 O_(4)^(2-) = 18.30 xx (1)/(1000 ) xx (0.02000)/(1) xx 5/2 = ( 0.0009150 mol` of ` C_2 O_(4)^(2-)`
`[ H_2 C_2 O_4] = 0.06100 M`