A solid mixture of sodium oxalate ( ) and potassium binoxalate-oxalic acid double salt ( ) required — Redox Reactions and Volumetric Analysis Chemistry Question
Question
A solid mixture of sodium oxalate ( $Na_2C_2O_4$ ) and potassium binoxalate-oxalic acid double salt ( $KHC_2O_4 \cdot H_2C_2O_4$ ) required equal volumes of $0.1 M$ $KMnO_4$ (in acidic medium) and $0.1 M$ $NaOH$ solution in two entirely separate titration experiments. What is the precise molar ratio of sodium oxalate to the double salt in the original mixture?
💡 Solution & Explanation
Let moles of $Na_2C_2O_4$ be $a$ and the double salt be $b$ . For $KMnO_4$ titration: $Na_2C_2O_4$ provides 1 oxalate group ( $n=2$ ), the double salt provides 2 oxalate groups ( $n=2+2=4$ ). Total redox equivalents = $2a + 4b = 0.1 \times V \times 5 = 0.5V$ . For $NaOH$ titration: $Na_2C_2O_4$ is an entirely neutral salt ( $n=0$ ). The double salt has exactly 3 ionizable acidic protons ( $n=3$ ). Total acid-base equivalents = $3b = 0.1 \times V \times 1 = 0.1V$ . By substituting $V = 30b$ into the redox equation: $2a + 4b = 0.5(30b) = 15b \Rightarrow 2a = 11b \Rightarrow a/b = 11/2 = 5.5 : 1$ .