The volume of 0.1 , M aqueous KOH solution re | vedclass

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Question

(A) $H_3PO_3$ is a dibasic acid,meaning it has two replaceable hydrogen atoms.The neutralization reaction is: $H_3PO_3 + 2KOH \rightarrow K_2HPO_3 + 2

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) $H_3PO_3$ is a dibasic acid,meaning it has two replaceable hydrogen atoms.The neutralization reaction is: $H_3PO_3 + 2KOH ightarrow K_2HPO_3 + 2H_2O$.Using the equivalence relation: $n_{acid} 	imes 	ext{basicity} = n_{base} 	imes 	ext{acidity}$.$M_1 	imes V_1 	imes n_f(acid) = M_2 	imes V_2 	imes n_f(base)$.$0.1 	imes 20 	imes 2 = 0.1 	imes V_2 	imes 1$.$V_2 = \frac{0.1 	imes 20 	imes 2}{0.1 	imes 1} = 40 \, mL$.

The volume of $0.1 \, M$ aqueous $KOH$ solution required for the complete neutralization of $20 \, mL$ of $0.1 \, M$ aqueous $H_3PO_3$ solution is .......... $mL$.

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