Question

1. Calculate the pH when 24.9 mL of 0.011 M HCl is added to 100.0 mL of the above buffer.

2. Calculate how many mL of 0.100 M NaOH are needed to neutralize completely 91.0 mL of 0.0600 M H₂SO₄ (forming Na₂SO₄ and water).

3.
Calculate the pH of 0.057 M phosphoric acid (H₃PO₄, a triprotic acid). K_a1 = 7.5 x 10^-3, K_a2 = 6.2 x 10^-8, and K_a3 = 4.8 x 10^-13.

Hint, if you are doing much work, you are making the problem harder than it needs to be.

Answer 1

3)   Answer:  When finding the pH of a solution of a polyprotic acid, almost all of the H3O+ produced comes from the first ionization step (Ka1). The H3O+ obtained from Ka2 and Ka3 are negligible.

Molarity . . . . . . H3PO4 + H2O ==> H3O+ + H2PO4-
Initial . . . . . . . . . 0.057 . . . . . . . . . . . .0 . . . . . . .0
Change . . . . . . . .-x . . . . . . . . . . . . . .x . . . . . . .x
Equilibrium . . . . 0.057-x . . . . . . . . . . . x . . . . . . .x

Ka1 = [H3O+][H2PO4-] / [H3PO4] = (x)(x) / ( 0.057-x) = 7.5 x 10^-3
x^2 / ( 0.057-x) = 7.5 x 10^-3 . . .Because x will not be negligible compared to 0.098, we can't drop it and we must solve a quadratic equation. (see reference below).

x^2 / ( 0.057-x) = 0.0075
x^2 = (0.0075)( 0.057-x)
x^2 = -0.0075x + 0.0004275
x^2 + 0.0075x - 0.0004275 = 0.. . . .using the quadratic formula we find that

x = 0.0046 M = [H3O+]
pH = -log [H3O+] = 2.34