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Chem 2 - Acid-Base Equilibria X: Buffers and the Henderson-Hasselbalch Equation

This document discusses buffers and the Henderson-Hasselbalch equation for calculating buffer pH. It defines buffers as solutions containing approximately equal concentrations of a weak acid and its conjugate base, or a weak base and its conjugate acid. The Henderson-Hasselbalch equation relates the pH of a buffer to its pKa value and the concentration ratio of the conjugate base to weak acid. The document provides an example calculation of the pH of an acidic buffer initially and after addition of hydrochloric acid, demonstrating how buffers resist changes in pH from small additions of acid or base.

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Acid-Base Equilibria (Pt. 10) Buffers and the Henderson- Hasselbalch Equation By Shawn P. Shields, Ph.D. This work is licensed by Dr. Shawn P. Shields-Maxwell under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
What is a Buffer? A buffer is a solution of approximately equal concentrations of either A weak acid and its conjugate base or A weak base and its conjugate acid
Acidic and Basic Buffers? Buffers made from a weak acid and its conjugate base are called “acidic buffers.” A buffers made from a weak base and its conjugate acid are called “basic buffers.”
Buffers Buffers resist changes in pH due to small additions of acid or base.
Henderson-Hasselbalch Equation Calculating the pH of a buffer solution 𝐇𝐀 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐥 ⇌ 𝐇 𝟑 𝐎+ 𝐚𝐪 + 𝐀− (𝐚𝐪) pH = pKa + log A− HA conjugate base weak acid
Henderson-Hasselbalch Equation Use this form of the equation for acidic and basic buffers. Always calculate pKa for the acid. pH = pKa + log A− HA base acid
Example: Calculating the pH of a Buffer Calculate the pH of a buffer made from 0.28 M HNO2 and 0.23 M NO2 . The Ka for HNO2 is 4.6  10-4.
Example: Calculating the pH of a Buffer Calculate the pH of a buffer made from 0.28 M HNO2 and 0.23 M NO2 . The Ka for HNO2 is 4.6  10-4. First, calculate pKa for the acid. pKa =  log Ka =  log (4.6  10-4) = 3.34
Example: Calculating the pH of a Buffer Calculate the pH of a buffer made from 0.28 M HNO2 and 0.23 M NO2 . The Ka for HNO2 is 4.6  10-4. Plug in pKa and the concentrations of the acid and conj base. pH = pKa + log A− HA pH = 3.34 + log 0.23 0.28 = 3.25 base acid
Example: Calculating the pH of a Buffer after the Addition of Acid or Base A buffer is prepared using 0.28 M HNO2 and 0.23 M NO2 . The pKa for HNO2 is 3.34. What is the pH of this buffer after the addition of 0.05 M HCl?
Example: Calculating the pH of a Buffer after the Addition of Acid or Base A buffer is prepared using 0.28 M HNO2 and 0.23 M NO2 . The pKa for HNO2 is 3.34. What is the pH of this buffer after the addition of 0.05 M HCl? HCl is a strong acid. It will react completely (and immediately) with the conjugate base (NO2 ) to form weak acid (HNO2).
Example: Calculating the pH of a Buffer after the Addition of Acid or Base A buffer is prepared using 0.28 M HNO2 and 0.23 M NO2 . The pKa for HNO2 is 3.34. What is the pH of this buffer after the addition of 0.05 M HCl? Since we added 0.05 M HCl… 0.05 M of the conjugate base (NO2 ) will react (be used) to form an additional 0.05 M weak acid (HNO2).
Example: Calculating the pH of a Buffer A buffer is prepared using 0.28 M HNO2 and 0.23 M NO2 . The pKa for HNO2 is 3.34. What is the pH of this buffer after the addition of 0.05 M HCl? Plug in pKa and the NEW concentrations of the acid and conj base. pH = 3.34 + log 0.23 − 0.05 0.28 + 0.05 = 3.08 0.18 0.33
Buffers Resist pH Changes Due to Addition of Acid or Base The pH of the buffer did not change much due to the addition of 0.05 M HCl. pH = 3.25 versus pH = 3.08 after addition. The buffer resisted the pH change due to the addition of acid.
Compare: Calculating the pH of a 0.05 M Strong Acid Solution What is the pH of water (not a buffer) after the addition of 0.05 M HCl pH =  log (0.05) = 1.30 The pH changed due to the addition of acid a lot (from pH = 7 to 1.30) !!!
Example problems will be posted separately. 