40 mL of 0.1 N KMnO_4 is equivalent to 30 | vedclass

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Question

(NONE) Step $1$: Determine the normality of $KHC_2O_4$ solution.
Using the equivalence principle,$N_1V_1 = N_2V_2$.
$0.1 \ N 	imes 40 \ mL = N_2

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(NONE) Step $1$: Determine the normality of $KHC_2O_4$ solution.
Using the equivalence principle,$N_1V_1 = N_2V_2$.
$0.1 \ N 	imes 40 \ mL = N_2 	imes 30 \ mL$.
$N_2 = \frac{4}{30} = \frac{2}{15} \ N$.
Step $2$: Titration of $KHC_2O_4$ with $KOH$.
$KHC_2O_4$ acts as an acid,reacting with $KOH$ in a $1:1$ molar ratio (or $1:1$ equivalent ratio).
$N_{acid} 	imes V_{acid} = N_{base} 	imes V_{base}$.
$\frac{2}{15} \ N 	imes 60 \ mL = 0.1 \ N 	imes V_{base}$.
$V_{base} = \frac{2 	imes 60}{15 	imes 0.1} = \frac{8}{0.1} = 80 \ mL$.

$40 \ mL$ of $0.1 \ N$ $KMnO_4$ is equivalent to $30 \ mL$ of $KHC_2O_4$ solution. How many $mL$ of $0.1 \ N$ $KOH$ are required to titrate $60 \ mL$ of the same $KHC_2O_4$ solution?

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