How to find the pH of the buffer solution when initiai pH=7 strength=0.1,volume=1L when 100ml of 1M phosphoric acid is added given(pka1=2.15,pka2=7.20,pka3=negligible) by the concept of ICE?

Answer 1

Phosphoric Acid H3PO4

100mL of 1M

pKa1 = 2.15, pKa2 = 7.2, pKa3 = negligible

H₃PO4 <--> H₂PO₄^- <--> HPO₄^2- <--> PO₄^3-

^pKa1 ^ pKa2

The first pKa is the most acidic (lowest pKa), so we start here.

H₃PO₄ <---> H₂PO₄^- + H⁺

Initial:

H₃PO₄ = 1M

H₂PO₄^- = 0

H⁺ = 0

Change:

H₃PO₄ - x

H₂PO₄^- + x

H⁺ + x

End

H₃PO₄ = 1M - x

H₂PO₄^- = x

H⁺ = x

pKa = -log(K_a)

-pKa = log(K_a)

K_a = 10^-pKa

K_a = ([H₂PO₄^-])*([H⁺]) / [H₃PO₄] = 10^-pKa

Substitute unknowns from ICE table into Ka equation.

K_a = x² / (1-x) = 10^-2.15

K_a = x² / (1-x) = 7.1x10^-3 = 0.0071

assume x is small (negligible difference on the denominator, cannot ignore on the numerator though)

K_a = x² / 1 = 0.0071

x = sqrt(0.0071) = 0.084

x = [H₂PO4^-] = [H⁺]

pH = -log([H⁺]) = -log(0.084) = 1.08

The second pKa, 7.2 has K_a = 10^-7.2 = 6.31x10^-8, several orders of magnitude below the first

Ka= 7.1x10^-3. The second pKa, 7.2, is very far from the pH contribution of the first pKa alone, 1.08, so there is very little deprotonation of H₂PO₄^- to contribute further.

Thus, 1.08 is the accepted pH.