Question

What is the pH of 0.41 M H₃PO₄? The K_a values for phosphoric acid are K_a1 = 7.5×10-3, K_a2 = 6.2×10-8, and K_a3 = 3.6×10-13.

Answer 1

Since Ka2 and Ka3 are very small as compared to Ka1
We will use only Ka1

H3PO4    -----> H +   H2PO4 -
0.41                      0           0       (initial)
0.41-x                  x           x        (at equilibrium)

Ka1 = [H+] [H2PO4-]/[H3PO4]
7.5*10^-3 = x*x / (0.41-x)

3.075*10^-3 - 7.5*10^-3 x = x^2
x^2 + 7.5*10^-3 x - 3.075*10^-3 = 0

solving for x, the positive value of x is 0.052

So,
[H+] = 0.052 M
pH = -log [H+]
       = -log (0.052)
       = 1.28
Answer: 1.28