# Enduring Understanding 6.C: Titration Curves

- When a solution of a weak acid HA is titrated with a strong base like NaOH, the
**titration curve**(graph of pH vs. NaOH added) looks like this: - When one half an equivalent of NaOH is added, the pH is equal to the pK
_{a}of the weak acid. - At the half-equivalence point, half the acid has reacted to form the conjugate base, so [HA] = [A
^{-}] - According to the Henderson Hasselbach equation:
- Since log [A
^{-}]/[HA] = 0, pH = pK_{a} - At the equivalence point, all the acid has reacted with the base, so the solution is effectively a solution of NaA.
- Sample Question 1: What is the pH of 50 mL of a 0.1M CH
_{3}COOH solution titrated to its equivalence point with 50 mL of an 0.1M NaOH? pKa of acetate = 4.75 - All the CH
_{3}COOH has been converted to CH_{3}COO^{-} - The concentration of CH
_{3}COO^{-}is (0.05L x 0.1 mol/L) / 0.10 L or 0.05 mol/L - The CH
_{3}COO^{-}will react with the water to form hydroxide: - We can use the K
_{b}of acetate to calculate the concentration of hydroxide present. - The pK
_{b}for acetate is (14-pK_{a}) or 9.25, so the K_{b}is 10^{-9.25}or 5.6 x 10^{-10} - [OH
^{-}] □ [CH_{3}COOH] - K
_{b}= 5.6 x 10^{-10}= [CH_{3}COOH] [OH^{-}] / [CH_{3}COO^{-}] = [OH^{-}]^{2}/ 0.05 mol/L - [OH
^{-}]^{2}= 2.8 x 10^{-11}, so [OH^{-}] = 5.3 x 10^{-6} - pOH= -log [OH
^{-}] = 5.27 - pH = (14-pOH) = 8.73
- Titration curves of polyprotic acids are more complex. Here is the titration curve of a triprotic acid:
- If the starting acid was H
_{3}PO_{4}, then: - Midpoint 1 would represent the point where [H
_{3}PO_{4}]=[H_{2}PO_{4}^{-}] - Midpoint 2 would represent the point where [H
_{2}PO_{4}^{-}]=[HPO_{4}^{2-}] - Midpoint 3 would represent the point where [HPO
_{4}^{2-}]=[PO_{4}^{3-}] - Equivalence point 1 represent the point where the [H
_{3}PO_{4}] was completely reacted with the base. - Equivalence point 2 represent the point where the [H
_{2}PO_{4}^{-}] was completely reacted with the base. - Equivalence point 3 represent the point where the [HPO
_{4}^{2-}] was completely reacted with the base.

_{a}+ log [A

^{-}]/[HA]

_{3}COO

^{-}+ H

_{2}O → OH

^{-}+ CH

_{3}COOH

Related Links:Chemistry Chemistry Quizzes AP Chemistry Notes Le Chatelier's Principle |

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