V. Buffers and Polyprotic Acids
VI-1. Buffers VI-2. Polyprotic Acids VI-3. Problem List VI-4. Buffer Problems VI-5. Polyprotic Acid Problems

Introduction

Polyprotic acids are acids that possess more than one acidic proton.

Example: H3PO4

H3PO4 (aq) <--> H2PO4-(aq) + H+(aq)        Ka1 = 7.11x10-3     pKa1 = 2.15

H2PO4-(aq) <--> HPO42-(aq) + H+(aq)        Ka2 = 6.34x10-8     pKa2 = 7.20

HPO42-(aq) <--> PO43-(aq) + H+(aq)        Ka3 = 4.20x10-13     pKa3 = 12.38


Sample Problem

What is the pH of the resulting solution when 50.0 mL of 0.700 M NaOH is mixed with 50.0 mL of 0.500 M H3PO4?

1. It is easiest to consider the initial reactions stepwise:

moles of H3PO4 = (0.0500 L)(0.500 M) = 0.0250 moles

moles of OH- = (0.0500 L)(0.700 M) = 0.0350 moles

0.0250 moles of H3PO4 neutralizes 0.0250 moles of OH-. We now have 0.0100 moles of OH- and 0.0250 moles of H2PO4-. The original amount of H3PO4 is completely consumed.

0.0100 moles of H2PO4- neutralizes the remaining OH-, leaving 0.0150 moles of H2PO4- and 0.0100 moles of HPO42-.

2. So the equilibrium is:     H2PO4-(aq) <--> HPO42-(aq) + H+(aq)        Ka2 = 6.34x10-8

and the equilibrium expression is:

        [H+][HPO42-]
Ka2  =  ------------
          [H2PO4-]

3. and 4. We have formed a buffer so steps 3. and 4. are not needed explicitly.

5. We can now calculate the equilibrium concentrations using the equilibrium constant expression:

        [H+][HPO42-]
Ka2  =  ------------
          [H2PO4-]

             [H+](0.0100 moles/0.100 L)
6.34x10-8  =  ------------------------
               (0.0150 moles/0.100 L)
Note that the volume appears in the numerator and denominator and cancels out.

[H+] = (6.34x10-8) * (0.0150) / (0.0100)

[H+] = 9.51x10-8

This problem asked for the pH of the solution.

pH = -log[H+]

pH = -log(9.51x10-8)

pH = 7.02


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