40 text{ mL} of x text{ M } KMnO_4 solution i | vedclass

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(A) In acidic medium,the balanced redox reaction is:$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$Using the equivalence princ

Vedclass · Accepted Answer

$0.04$

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(A) In acidic medium,the balanced redox reaction is:$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ ightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$Using the equivalence principle,the number of equivalents of $KMnO_4$ equals the number of equivalents of oxalic acid:$n_1 	imes M_1 	imes V_1 = n_2 	imes M_2 	imes V_2$For $KMnO_4$,the change in oxidation state of $Mn$ is from $+7$ to $+2$,so $n_1 = 5$.For oxalic acid $(H_2C_2O_4)$,the change in oxidation state of $C$ is from $+3$ to $+4$ for two carbon atoms,so $n_2 = 2$.Substituting the given values:$5 	imes x 	imes 40 = 2 	imes 0.02 	imes 200$$200x = 8$$x = \frac{8}{200} = 0.04 	ext{ M}$

$40 \text{ mL}$ of $x \text{ M } KMnO_4$ solution is required to react completely with $200 \text{ mL}$ of $0.02 \text{ M}$ oxalic acid solution in acidic medium. The value of $x$ is:

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