Question

\[\ce{P4O6}\] reacts with water according to equation \[\ce{P4O6 + 6H2O -> 4H3PO3}\]. Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of \[\ce{P4O6}\] in \[\ce{H2O}\].

Answer 1

\[\ce{P4O6 + 6H2O -> 4H3PO3}\]

\[\ce{H3PO3 + 2NaOH -> Na2 HPO3 + 2H2O × 4}\]  ....(Neutralisation reaction)

\[\ce{P4O6 + 8NaOH -> 4Na2 HPO4 + 2H2O}\]

1 mol  8 mol

Product formed by 1 mol of \[\ce{P4O6}\] is neutralised by 8 mols of \[\ce{NaOH}\]

∴ Product formed by `1.1/220` mol of \[\ce{P4O6}\] will be neutralised by `1.1/220 xx 8` mol of \[\ce{NaOH}\]

Molarity of \[\ce{NaOH}\] solution is 0.1 M

⇒ 0.1 mol \[\ce{NaOH}\] is present in 1 L solution

∴ `1.1/220 xx 8` mol \[\ce{NaOH}\] is present in `(1.1 xx 8)/(220 xx 0.1)`L

= `88/220`L

= `4/10`L

= 0.4 L

= 400 mL of \[\ce{NaOH}\] solution.