A titration mixture having pottasium oxalate and pottasium hydrogen oxalate!

Question

I came across this question, this is how it goes-

A solution is prepared by dissolving a solid mixture of $\ce{K2C2O4}$ and $\ce{KHC2O4}$. A $\pu{10 ml}$ portion of this solution required $\pu{10 ml, 0.05 M \ce{KOH}}$ solution for titration reaction. In a seperate analysis $\pu{10 ml}$ of the same stock solution required $\pu{10 ml, 0.06 M}$ acidified $\ce{KMnO4}$ solution for titration.

I have to find the ratio of $\ce{K2C2O4}$ to $\ce{KHC2O4}$ in the initial reaction mixture.

I know that(I think), 1 mole of $\ce{KHC2O4}$ would require 1 mole of of $\ce{KOH}$ for neutralization. While $\ce{K2C2O4}$ would not react with $\ce{KOH}$ (please correct me if I am wrong).

Also, $\pu{31.8 g}$ of $\ce{KMnO4}$ would oxidise $\pu{63.5 g}$ of $\ce{K2C2O4}$. But when I don't know how (differently) it shall oxidise $\ce{KHC2O4}$.

Answer 1

You are right, $1$ mole of $\ce{KHC2O4}$ reacts with $1$ mole of $\ce{KOH}$. From that first piece of information in your exercise, you can find out that, in $\pu{10mL}$ of solution, you have $\pu{0.05 * 0.01 = 5 * 10^{-4}}$ moles of $\ce{KHC2O4}$. The next piece of information is meant to let you how much substance there is in the solution. $10$ moles of both substances react with $4$ moles of acidified $\ce{KMnO4}$. Which means the sum of the quantities of both substances in that solution is $\pu{(0.06 * 0.01 * 10) / 4 = 15 * 10^-4}$. So you got $\pu{a = 5 * 10^{-4}}$ and $\pu{a + b = 15 * 10^{-4}}$. Do the math.

Please tell me if I was clear enough.