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

Explanation

`P_(4)O_(6) + 6H_(2)O to 4H_(3)PO_(3)`
`([H_(3)PO_(3) + 2NaOH " "Na_(2)HPO_(3) + 2H_(2)O] xx4)/(P_(4)O_(6)(1 "mol") + 8 NaOH( 8" mol") to 4Na_(2)HPO_(3) + 2H_(2)O)`
Prouduct formed by 1 mol `P_(4)O_(6)` is be neutralised by `(1.1)/(220) xx 8 "mol" ` of NaOH. Again molarity of NaOH soluiton is 0.1(M).
Therefor , 0.1 mol NaOH is present in 1 L solution .
`therefore (1.1)/(220) xx 8 mol NaOH ` is present in `(1.1 xx 8)/(220 xx 0.1)L`
` = 0.4L = 400m L` NaOH solution .