P 4 O 6 reacts with water according to equation P 4 O 6 + 6H 2 O → 4H 3 PO 3 . Calculate the volume of 0 .1 M NaOH (in mL ) solution required to neutralise the acid formed by dissolving 1 .1 g of P 4 O 6 in H 2 O?

Step-by-step solution

P 4 O 6 + 6H 2 O → 4H 3 PO 3 ......(i) Neutralisation : H 3 PO 3 + 2NaOH → Na 2 HPO 3 + 2H 2 O × 4 .......(ii) Adding Eqs. (i) and (ii) P 4 O 6 + 8 NaOH ⟶ 4 Na 2 HPO 3 + 2 H 2 O 1 mol 8 mol .....(iii) Number of moles of P 4 O 6 , n = m M = 1 . 1 2 2 0 = 1 2 0 0 mol (Molar mass P 4 O 6 = ( 4 × 31 ) + ( 6 × 16 ) = 220 ∵ Product formed by 1 mole of P 4 O 6 is neutralised by 8 moles NaOH ∴ Product formed by 1 2 0 0 moles of P 4 O 6 will be neutralised by NaOH = 8 × 1 2 0 0 = 8 2 0 0 moles NaOH Given, Molarity of NaOH =