V. Buffers and Polyprotic Acids | ||||
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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: H_{3}PO_{4}
H_{3}PO_{4 (aq)} H_{2}PO_{4}^{-}_{(aq)} + H^{+}_{(aq)} K_{a1} = 7.11x10^{-3} pK_{a1} = 2.15
H_{2}PO_{4}^{-}_{(aq)} HPO_{4}^{2-}_{(aq)} + H^{+}_{(aq)} K_{a2} = 6.34x10^{-8} pK_{a2} = 7.20
HPO_{4}^{2-}_{(aq)} PO_{4}^{3-}_{(aq)} + H^{+}_{(aq)} K_{a3} = 4.20x10^{-13} pK_{a3} = 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 H_{3}PO_{4}?
1. It is easiest to consider the initial reactions stepwise:
moles of H_{3}PO_{4} = (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 H_{3}PO_{4} neutralizes 0.0250 moles of OH^{-}. We now have 0.0100 moles of OH^{-} and 0.0250 moles of H_{2}PO_{4}^{-}. The original amount of H_{3}PO_{4} is completely consumed.
0.0100 moles of H_{2}PO_{4}^{-} neutralizes the remaining OH^{-}, leaving 0.0150 moles of H_{2}PO_{4}^{-} and 0.0100 moles of HPO_{4}^{2-}.
2. So the equilibrium is: H_{2}PO_{4}^{-}_{(aq)} HPO_{4}^{2-}_{(aq)} + H^{+}_{(aq)} K_{a2} = 6.34x10^{-8}
and the equilibrium expression is:
[H^{+}][HPO_{4}^{2-}] K_{a2} = ------------ [H_{2}PO_{4}^{-}]
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^{+}][HPO_{4}^{2-}] K_{a2} = ------------ [H_{2}PO_{4}^{-}] [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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