H2PO4-(aq) + H+(aq) Ka1 = 7.11x10-3 pKa1 = 2.15
| 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 |
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
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 |
| Equilibrium Practice Problems | |
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