Only 2 mL of solution of unknown molarity is required to reach the end point of a titration of 20 mL of oxalic acid (2 M) in acidic medium. The molarity of solution should be ____________ M.

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Solution

Q.
Only 2 mL of ${KMnO}_{4}$ solution of unknown molarity is required to reach the end point of a titration of 20 mL of oxalic acid (2 M) in acidic medium. The molarity of ${KMnO}_{4}$ solution should be ____________ M. [2024]

Ans.
(8)

$K \overset{+ 7}{Mn} O_{4} + H_{2} \overset{+ 3}{C_{2}} O_{4} \to {Mn}^{2 +} + \overset{+ 4}{C} O_{2}$ (unbalanced)

$(n f a c t o r = 5) (n f a c t o r = 1 \times 2 = 2)$

$n factor of {KMnO}_{4}$

$n_{{KMnO}_{4}} = change in O . N . of Mn \times no . of Mn atoms in 1 molecule of {KMnO}_{4}$

$= 5 \times 1 = 5$

$n factor of H_{2} C_{2} O_{4}$

$n H_{2} C_{2} O_{4}$ = change in O. N. of C $\times$ no. of C atoms in 1 molecule of $H_{2} C_{2} O_{4}$

$= 1 \times 2 = 2$

By law of equivalence:

Equivalents of ${KMnO}_{4}$ = Equivalents of $H_{2} C_{2} O_{4}$

$n factor of {KMnO}_{4} \times$ moles of ${KMnO}_{4}$ = $n factor of H_{2} C_{2} O_{4}$ $\times$ moles of $H_{2} C_{2} O_{4}$

$5 \times$ millimoles of ${KMnO}_{4} = 2 \times$ millimoles of $H_{2} C_{2} O_{4}$

$5 \times M_{{KMnO}_{4}} \times V_{{KMnO}_{4}} (in mL) = 2 \times M_{H_{2} C_{2} O_{4}} \times V_{H_{2} C_{2} O_{4}} (in mL)$

$5 \times M_{{KMnO}_{4}} \times 2 mL = 2 \times 2 M \times 2 mL$

$M_{{KMnO}_{4}} = 8 M$

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