For the diprotic weak acid H2A, Ka 3.7 x 106 and Ka25.3 x10.9 What is the pH of

Question

For the diprotic weak acid H2A, Ka 3.7 x 106 and Ka25.3 x10.9 What is the pH of a 0.0700 M solution of H2A? What are the equilibrium concentrations of H2A and A-in this solution? STRATEGY: Start by making a table for the molar concentrations after the first dissociation step H,Ala q) H+(aq) + HA-(aq) Initial:0.0700 Change: -x press Ka1 in terms of x, and then show the simplified expression (when x is assumed to be negligible compared to to the initial concentration) Also, give the numerical value of x using the appropriate approximation. What is the pH of the solution you stopped here? Number K. Number Complete Simplified

Explanation / Answer

SOLUTION:

H2A (aq) <---------> H+(aq) + HA-(aq)

I 0.0700M 0 0

C -x x x

E 0.0700M-x x x

Ka1 = [x][x] / [0.07 - x] = x2 / [0.07 - x] Simplified expression = Ka1 = x2 / 0.07   ; Because x << 0.07

3.7 X 10-6 [0.07 - x] = x2

x2 + 3.7 X 10-6  x - 2.59 X 10-7 = 0

This is a quadraitic equation its solution gives value of x.

x = 0.005M

It means HA- = H+ = 0.0005M

H2A = 0.0700 - x = 0.0700 - 0.0005 = 0.0695M

Now HA- will dissociate as:

HA- <----------> A- + H+

I 0.0005M 0 0

C -x x x

E 0.0005-x x x

Ka2 = x2 / [0.0005 - x] Simplified expresssion = Ka2 = x2 / 0.0005 ; x << 0.0005

5.3 X 10-9 = x2 / [0.0005 - x]  

5.3 X 10-9 [0.0005 - x] = x2

x2 + 5.3 X 10-9 x - 2.6 X 10-12 = 0

solving this quadratic equation gives x.

x = 1.6 X 10-6M

It means A- = 1.6 X 10-6M

pH = -log[H+]

[H+] = 0.0005M + 1.6 X 10-6M = 0.0005M

pH = -log[0.0005] = 3.3

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