Phosphorus pentoxide reacts with water to form phosphoric acid according to the following equation - HSC - SSCE Chemistry - Question 10 - 2006 - Paper 1

According to the balanced equation:

$P_2O_5 + 3H_2O \rightarrow 2H_3PO_4$

1 mole of P₂O₅ produces 2 moles of H₃PO₄. Therefore, the moles of H₃PO₄ produced is:

$\text{Moles of } H_3PO_4 = 2 \times \text{Moles of } P_2O_5 = 2 \times 0.0100 \, mol = 0.0200 \, mol$

The reaction between H₃PO₄ and NaOH is:

$H_3PO_4 + 3NaOH \rightarrow Na_3PO_4 + 3H_2O$

1 mole of H₃PO₄ reacts with 3 moles of NaOH, thus:

$\text{Moles of } NaOH = 3 \times \text{Moles of } H_3PO_4 = 3 \times 0.0200 \, mol = 0.0600 \, mol$

Using the concentration formula:

$C = \frac{n}{V}$

we can rearrange to find volume:

$V = \frac{n}{C}$

Where:

• $n = 0.0600 \, mol$ (moles of NaOH)
• $C = 0.30 \, mol/L$

Plugging in the values:

$V = \frac{0.0600 \, mol}{0.30 \, mol/L} = 0.200 \, L$

Therefore, the volume of 0.30 mol L$^{-1}$ sodium hydroxide required to neutralise the phosphoric acid produced is:

0.20 L (Option C)