`P_4O_6` reacts with water according to equation `P_4O_6 + 6H_2O to 4H_3PO_3`. Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of `P_(4)O_(6)` in `H_2O` .

Explanation

Mol. Mass of `P_(4)O_(6) = 220` g/mol
Moles of `P_(4)O_(6)` dissolved in `H_(2)O = (1.1)/220`
`underset(1 mol)(P_(4)O_(6)) + underset(6 "mol")(6H_(2)O) to underset(4 "mol")(4H_(3)PO_(3))`
`therefore` Moles of `H_(3)PO_(3)` produced `=4.4/(220)`
`underset(1 mol)(H_(3)PO_(3)) + 2NaOH to Na_(2)HPO_(3) + 2H_(2)O`
`therefore (4.4)/(220) mol`
Moles of NaOH consumed `=(2 xx 4.4)/220 = 8.8/(220)`
`therefore` Volume of NaOH required
`=("Moles of NaOH" xx 1000)/("Molarity of NaOH")`
`=(8.8 xx 1000)/(220 xx 0.1)`
= 400 mL
`V_(NaOH) = 400 mL`