`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 for the reactions involved are :
`{:(" "P_(4)O_(6) + 6H_(2)O rarr 4 H_(3)PO_(4)" "("Hydrolysis reaction")),(ul(H_(3)PO_(3) + 2 NaOH rarr [4 Na_(2)HPO_(3)+ 2H_(2)O]xx 4) " "("Neutralisation reaction")),(" "underset("= 220 g")underset(4 xx 31 + 6 xx 16)(P_(4)O_(6))+underset("= 320 g")underset(8 xx 40)(8 NaOH) rarr 4 Na_(2)HPO_(3)+2H_(2)O" "("Overall reaction")):}`
Now 220 g of `P_(4)O_(6)` require NaOH for neutralization = 320 g
`therefore` 1.1 g of `P_(4)O_(6)` will require NaOH `= (320)/(220) xx 1.1 = 1.6 g`
Now 1000 mL of 0.1 MHCl contain NaOH `= 40 xx 0.1 = 4 g`
In other words, 4 g of NaOH are present in 0.1 M NaOH = 1000 mL
`therefore` 1.6 g of NaOH will be present in 0.1 M NaOH `=(1000)/(4) xx 1.6 = 400 mL = 0.4 L`