`P_(4)O_(6)` reacts with water according to equation `P_(4)O_(6)to4H_(3)PO_(3)`.
Calculate the volume of `0.1MNaOH` solution required to neutralise the acid formed by dissolving 1.1g of `P_(4)O_(6)` in `H_(2)O`.

Explanation

The chemical equations involved in the reactions are :
`P_(4)O_(6) +6H_(2)O to 4H_(3)PO_(3)`
`(H_(3)PO_(3)+2NaOH to Na_(2)HPO_(4)+2H_(2)Oxx4)/(P_(4)O_(6)+8NaOH to 4 Na_(2)HPO_(4)+2H_(2)O)`
No. of moles of `P_(4)O_(6)` in 1.1 g `=("Mass of" P_(4)O_(6))/("Molar mass")=((1.1g))/((220 g "mol"^(-1)))=0.005` mol
1 mole of `P_(4)O_(6)` require NaOH=8 mol.
0.005 mole of `P_(4)O_(6)` requir `NaOH=8xx0.005=0.04` mol
Molarity of NaOH solution `=("No of moles of NaOH")/("Volume of solution")`
0.1 M `(0.1 "mol L"^(-1))=((0.04 mol))/("volume of solution)`
Volume of NaOH solution `=((0.04 mol))/((0.1 mol L^(-1)))=0.4 L=400 mL`