`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

`P_(4)O_(6)+6H_(2)Oto4H_(3)PO_(3)` . . .(i)
Neutralisation
`H_(3)PO_(3)+2NaOHtoNa_(2)HPO_(3)+2H_(2)]xx4` . . .(ii)
Adding Eqs. (i) and (ii)
`underset("1mol")(P_(4)O_(6))+underset("8mol")(NaOH)to4Na_(2)HPO_(3)+2H_(2)O` . . . (iii)
Number of moles of `P_(4)O_(6)`,
`n=(m)/(M)=(1.1)/(220)=(1)/(200)mol`
(Molar mass of `P_(4)O_(6)=(4xx31)+(6xx16)=220`
`:'` Product formed by 1 mole of `P_(4)O_(6)` is neutralised by 8 moles `NaOH`
`:.` Product formed by `(1)/(200)` moles of `P_(4)O_(6)` will be neutralised by `NaOH`
`=8xx(1)/(200)=(8)/(200)"mole" NaOH`
Given, Molarty of `NaOH=0.1M=0.1mol//L`
Molarty `=("Number of moles")/("Volume in litres")`
Volume `=("Number of moles")/("Molarity")=(8)/(200)xx(1)/(0.1)=0.4L or 400 mL`
`:. 400 mL NaOH` is required.