`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_(4)" (Hydrolysis reaction)"),(H_(3)PO_(3)+2NaOHtoNa_(2)HPO_(3)+2H_(2)O"]"xx4 " (Neutrolisation reaction)"),(overline(underset(=220g)underset(4xx31+6xx16)(P_(4)O_(6))+underset(=320 g)underset(8xx40)(8NaOH )to 4Na_(2)HPO_(3)+2H_(2)O)" (Overall reaction)"):}`
Now 220 g of `P_(4)O_(6)` require NaOH for neutralization =320 g
`:.` 1.1 g of `P_(4)O_(6)` will require `NaOH=(320)/(220)xx1.1=1.6g`
Now 1000 mL of 0.1 MHCl contain `NaOH=40xx0.1=4g`
In other words, 4 g of NaOH are present in 0.1 M NaOH `=(1000)/(4)xx1.6=400mL=0.4L`