In: Chemistry
A 17.2−g 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=
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