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The concentration of oxalic acid is ' $X$ ' ${mollit}^{- 1}$ . $40 ml$ of this solution reacts with $16 ml$ of 0.05 $M$ acidified ${KMnO}_{4}$ . What is the $pH$ of ' $X$ ' $M$ oxalic acid solution? (Asssume that oxalic acid dissociates completely)

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1.3

1.699

answer is C.

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Detailed Solution

$2 {KMnO}_{4} + 3 H_{2} {SO}_{4} + 5 {(C O O H)}_{2} \to \to K_{2} {SO}_{4} + 2 {MnSO}_{4} + 8 H_{2} O + 10 {CO}_{2}$

$\frac{M_{1} V_{1}}{n_{1}} ({KMnO}_{4}) = \frac{M_{2} V_{2}}{n_{2}} {(\begin{matrix} COOH \end{matrix})}_{2}$

$\frac{0.5 \times 16}{2} = \frac{x \times 40}{5}$

$x = 0.05 M$

$[H^{+}] = 2 \times 0.05 = 0.1 M$

$pH = - \log (H^{+} = - \log (0.1) = 1$

Hence the correct answer is (C)1

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