Question

A 24.7-g sample of white phosphorus was burned in excess of oxygen. The product was dissolved in enough water to make 549 mL of solution. Calculate the pH of the solution at 25 C

Answer 1

Solution :-

White phosphorous = P4

Lets first calculate moles of P4

Moles of P4 = mass / molar mass

                     = 24.7 g / 123.895 g per mol

                     = 0.199 mol P4

Total moles of P = 0.199 mol * 4 = 0.796 mol P

Each mole produce H3PO4 acid when the oxidation product is dissolved in water

So moles of H3PO4 that can be produced = 0.796 mol

Molarity = moles / liter

               = 0.796 mol / 0.549 L

              = 1.5 M

Now using the Ka values of the H3PO4 lets calculate the concentration of the H+

H3PO4 + H2O ----- > H3O+ + H2PO4

1.5 M                               0              0

-x                                       +x         +x

1.5-x                                    x          x

Ka1=[H3O+][H2PO4^-]/[H3PO4]

7.2*10^-3 = [x][x]/[1.5-x]

7.2*10^-3 * (1.5-x) =x^2

Solving for x we get

x=0.100

So concentration of the H3O+ and H2PO4^2- = 0.100 M after first dissociation

Now lets write the second dissociation

H2PO4- + H2O   ------ > H3O+   + HPO4^2-

0.100                                  0               0

-x                                      +x            +x

0.100-x                             x              x

Ka2= [H3O+][HPO4^2-]/[H2PO4^-]

6.3*10^-8 = [x][x]/[0.100-x]

Ka2 is small therefore we can neglect the x from denominator

Then we get

6.3*10^-8=[x][x]/[0.100]

6.3*10^-8 *0.100 = x^2

6.3*10^-9 = x^2

Taking square root of both sides we get

7.94*10^-4 = x

So the H3O+ = 7.94*106-5 M

Ka3 is very small therefore we neglect the third dissociation

So the total [H3O+] = 0.100 + 7.94*10^-5 = 0.100 M

Now lets calculate the pH

pH=-log [H3O+]

pH= -log [0.100]

pH= 1

So the pH of the solution is 1