Question

Calculate the pH at 25 C of a 0.75 M aqueous solution of phosphoric acid (H3PO4). (Ka1, Ka2, and Ka3 for phosphoric acid are 7.5*10^-3, 6.25*10^-8, and 4.8*10^-13, respectively.)

Answer 1

M = 0.75 M of H3A

H3A <-> H+ H2A^-

H2A <-> H+ HA^-2

HA- <-> H+ A^-3

This will be formed...

We need the [H+] concentration

From the first equilbrium

Ka1 = [H2A^-][H+] / [H3A]

Ka2 = [HA^2-][H+] /[H2A]

Ka3 = [A^-3 ][H+]/ [HA-]

There will be three H+ to account, the Ka1, Ka2 and form Ka3

Let us calculate first the ammount of Ka1 H+ ions

NOTE: [H2A-] = [H+] = x since for 1 mol of [H+] you will have always another mol of HA-

[H3A] = 0.75 - x

Where x is the acid already in solution

Ka = 7.5*10^-3

Substitute in Ka1

Ka1 = [H2A-][H+] /[H3A]

7.5*10^-3 = x*x / (0.75 - x )

Solve for x

x =0.0713 and -0.0788

ignore the negative solution since there are no negative concentrations

x = [H+] = [H2A-] = 0.0713

Please remember this number, we will use it at the end when calculating total H+ moles

HINT: we could ignore the second equilibrium since Ka2 is too low compared to the Ka1. That is, the 95% or more of the acid is given by the first ionization

But lets do the other Ka2 equilibrium

Assume, once again

[HA^-2 ]= [H+] = y NOTE that these H+ ions are from the SECOND ionization and cannot be compared with the first ionization, thats why I'm using x and y to denote difference

[HA-] = x-y

[HA-] = x will be before the second dissociation, after the dissociation you need to account for the lost acid which is denoted as -y

Ka2 = 6.25*10^-8

Ka2 = [HA^-2 ][H+]/ [H2A-]

Substitute all data

6.25*10^-8 = y*y / (x-y)

6.25*10^-8= y² / (0.0713 -y)

Sovle for y

y = 6.67*10^-5 and -6.67*10^-5

Ignore negative values, no negative concentrations exist

Then

[H+] = y = 6.67*10^-5

Time to add [H+] of first ionization +[H+] of second ionization

that is x+y = 0.0713+ 6.67*10^-5 = 0.07113667

As you can see, the effect of the second ion is not enough to change the value of x significantively... I will not continue with the third ionization H+ since it is 10^-3 (vs 10^-8 and 10^-3) Doing this will surely not affect the pH

pH = -log[H+] = -log(0.07113667) = 1.1479

pH = 1.15