AP Chemistry 8.9 Henderson-Hasselbalch Equation Study Notes - New Syllabus Effective fall 2024
AP Chemistry 8.9 Henderson-Hasselbalch Equation Study Notes- New syllabus
AP Chemistry 8.9 Henderson-Hasselbalch Equation Study Notes – AP Chemistry – per latest AP Chemistry Syllabus.
LEARNING OBJECTIVE
Identify the pH of a buffer solution based on the identityand concentrations of theconjugate acid-base pair used to create the buffer.
Key Concepts:
- The Henderson-Hasselbalch Equation
Determining Buffer pH
The pH of a buffer solution depends on two key factors:
- The pKₐ of the weak acid (a measure of its acid strength).
- The ratio of the concentrations of the conjugate base (\( \mathrm{[A^-]} \)) and the acid (\( \mathrm{[HA]} \)) in the buffer.
This relationship is quantitatively expressed by the Henderson–Hasselbalch equation:
\( \mathrm{pH = pK_a + \log\!\left(\dfrac{[A^-]}{[HA]}\right)} \)
Key Understanding:
This equation is derived from the equilibrium expression for the weak acid dissociation:
\( \mathrm{HA \rightleftharpoons H^+ + A^-} \)
\( \mathrm{K_a = \dfrac{[H^+][A^-]}{[HA]}} \)
Taking the negative log of both sides gives the Henderson–Hasselbalch form.
When \( \mathrm{[A^-] = [HA]} \), the logarithmic term equals zero, and thus:
\( \mathrm{pH = pK_a} \)
This is the point of maximum buffer capacity.
- When small amounts of acid or base are added, the ratio \( \mathrm{[A^-]/[HA]} \) changes slightly, but pH remains nearly constant.
Key Idea: A buffer resists pH change because the conjugate acid neutralizes added base, and the conjugate base neutralizes added acid. The Henderson–Hasselbalch equation allows quick pH determination from the composition of the buffer.
Example :
Calculate the pH of a buffer containing 0.25 M acetic acid (\( \mathrm{CH_3COOH} \)) and 0.30 M sodium acetate (\( \mathrm{CH_3COONa} \)). \( \mathrm{pK_a(CH_3COOH) = 4.76} \).
▶️ Answer / Explanation
Step 1: Identify acid and conjugate base.
- \( \mathrm{HA = CH_3COOH} \)
- \( \mathrm{A^- = CH_3COO^-} \)
Step 2: Apply the Henderson–Hasselbalch equation:
\( \mathrm{pH = pK_a + \log\!\left(\dfrac{[A^-]}{[HA]}\right)} \)
Step 3: Substitute known values:
\( \mathrm{pH = 4.76 + \log\!\left(\dfrac{0.30}{0.25}\right)} \)
Step 4: Calculate:
\( \mathrm{pH = 4.76 + \log(1.20)} = 4.76 + 0.079 = 4.84 \)
Final Answer: \( \mathrm{pH = 4.84} \)
The pH is slightly greater than the pKₐ because the buffer contains slightly more conjugate base than acid.
Example:
Determine the pH of a buffer made from 0.20 M \( \mathrm{NH_3} \) and 0.25 M \( \mathrm{NH_4Cl} \). The \( \mathrm{K_b} \) for ammonia is \( 1.8 \times 10^{-5} \).
▶️ Answer / Explanation
Step 1: Convert \( \mathrm{K_b} \) to \( \mathrm{pK_a} \) of the conjugate acid:
\( \mathrm{pK_a = 14 – pK_b = 14 – 4.74 = 9.26} \)
Step 2: Apply the Henderson–Hasselbalch equation:
\( \mathrm{pH = pK_a + \log\!\left(\dfrac{[base]}{[acid]}\right)} \)
Step 3: Substitute values:
\( \mathrm{pH = 9.26 + \log\!\left(\dfrac{0.20}{0.25}\right)} \)
Step 4: Calculate:
\( \mathrm{pH = 9.26 + \log(0.80)} = 9.26 – 0.097 = 9.16} \)
Final Answer: \( \mathrm{pH = 9.16} \)
The pH is slightly lower than the pKₐ because the buffer contains more conjugate acid (\( \mathrm{NH_4^+} \)) than base (\( \mathrm{NH_3} \)).