What volume (in mL) of 0.1 M KOH solution i | vedclass

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

Vedclass · Accepted Answer

$40$

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(C) Phosphorous acid $(H_3PO_3)$ 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 stoichiometry of the reaction,$1 \ mole$ of $H_3PO_3$ reacts with $2 \ moles$ of $KOH$.Given: $M_1 = 0.1 \ M$,$V_1 = 20 \ mL$,$n_1 = 1$ (for $H_3PO_3$);$M_2 = 0.1 \ M$,$V_2 = ?$,$n_2 = 2$ (for $KOH$).Using the formula $M_1V_1 / n_1 = M_2V_2 / n_2$:$(0.1 	imes 20) / 1 = (0.1 	imes V_2) / 2$.$2 = 0.05 	imes V_2$.$V_2 = 2 / 0.05 = 40 \ mL$.

What volume (in $mL$) of $0.1 \ M \ KOH$ solution is required for the complete neutralization of $20 \ mL$ of $0.1 \ M$ phosphorous acid $(H_3PO_3)$?

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