JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Redox Reactions and Electrochemistry

10.0 ml of

0.05 M KMnO_{4}

solution was consumed in a titration with 10.0 mL of given oxalic acid dehydrate solution. The strength of given oxalic acid solution is ____

\times 1 0^{- 2} g / L

(Round off to the Nearest Integer).

Answer: 1575

Solution

In acidic medium permanganate is reduced to

M n^{2 +}

, a five-electron change, while oxalate is oxidised to carbon dioxide, a two-electron change:

2 Mn O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

Equating milliequivalents at the end point,

10.0 \times 0.05 \times 5 = 10.0 \times M \times 2 \Rightarrow M = 0.125 mol L^{- 1}

Oxalic acid dihydrate,

H_{2} C_{2} O_{4} \cdot 2 H_{2} O

, has a molar mass of

126

g mol

^{- 1}

, so the strength of the solution is

0.125 \times 126 = 15.75

g L

^{- 1}

, that is

1575 \times 1 0^{- 2}

g/L.

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer