Calculate the volume of 0.1 M NaOH solution required to neutralise the solution produced dissolving 1.1 g of `P_(4)O_(6)` in water.

Explanation

Step I. Calculation of the mass of phosphorus acid formed.
`underset(=220g)underset(31xx4+16xx6)(P_(4)O_(6))" "+3H_(2)O to " " underset(=3xx82=246 g)underset(3[3xx1+31+3xx16])(3H_(3)PO_(3))`
220 g og `P_(4)O_(6)` form phosphorus acid =246g
1.1 g of `P_(4)O_(6)` form phosphorus acid`=((246g))/((220 g))xx(1.1g)=1.23g`
Step II. Calculation of mass of NaOH required for neutralisation.
`H_(3)PO_(3)` is a dibasic acid and reacts with NaOH as follows :
`H_(3)PO_(3) " "+ 2NaOH to Na_(2)HPO_(3)+2H_(2)O`
`3+31+48=82g " " 2xx40=80 g`
82 g of `H_(3)PO_(3)` require NaOH=80g
1.23 g of `H_(3)PO_(3)` require `NaOH=((80 g))/((82 g))xx(1.23g)=1.2g`
Step III. Calculation of volume of NaOH solution required.
`"Molarity of NaOH"=(("Mass of NaOH")/("Molar mass"))/("Volume of solution in litres")`
`(0.1 "mol" L^(-1))=(((1.2g))/((40 g " mol"^(-1))))/(V)`
`V=((1.2g))/((40g" mol"^(-1))xx(0.1" mol" L^(-1)))=0.3 L=300 mL`