A 172g sample of white phosphorus was burned in excess of ox

A 17.2g sample of white phosphorus was burned in excess of oxygen. The product was dissolved in enough water to make 567 mL of solution. Calculate the pH of the solution at 25°C.

pH=

Solution

the reaction of white Phosphorous with oxygen which is in excess is

4P +5O2------->2P2O5

4 moles of P reacts with exces oxygen to produce 2 moles of P2O5

moles =mass/molar mass =17.2/31=0.55

4 moles of P gives 2 mole of P2O5

0.55 moles of P gives 0.55/2=0.275 moles of P2O5.

Volume of solution=567 ml=567/1000=0.567 l

concentration of P2O5= Moles/volume in l =0.275/0.567 =0.485M

P2O5 + 3 H2O 2 H3PO4

one mole of P2O5 gives 2 moles of H3PO4

0.485 M gives 2*0.485 =0.97 M H3PO4.

H3PO4 is triprotic acid and it undergoes ionization as

H3PO4 + H2O--------->H2PO4- + H3O+ , PKa1= 2.15 (1), H2PO4-+ H2O ------->H3O++ HPO4-2 , pKa2= 7.2 (2) and HPO4-2+ H2O------->PO4-2+ H3O+, pKa3= 12.15 (3)

of all the three Ka is 1st one is strong so contriburion from Eq.2 and 3 can be neglected

hence let x= drop in concentration of H3PO4 to reach equilibrium

at equilibrium [H2PO4-]= x= [H3O+] , [H3PO4]= 0.97-x

hence x²/(0.97-x)= Ka1= 10^(-2.15)= 0.0071, when solved using excel, x=0.0796

pH= -log (0.0796)= 1.1