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pH Buffering Solutions: Henderson-Hasselbalch Equation Explained

Learn how buffered solutions resist changes in pH when acid or base is added. Master the Henderson-Hasselbalch equation with examples and calculations. Understand the chemistry of buffer systems and how to maintain desired pH levels effectively.

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pH Buffering Solutions: Henderson-Hasselbalch Equation Explained

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  1. Buffered Solutions - A solution that can resist a change in pH when an acid, H+, or base, OH-, is added

  2. Typical Buffers • Usually acid/c.base or c.acid/base Examples: HC2H3O2/NaC2H3O2 or NH4Cl/NH3 Buffers can be made to achieve any pH.

  3. Calculate the pH of a 0.50 M HC2H3O2/0.50 M NaC2H3O2 solution. Ka = 1.8 x 10 -5 HC2H3O2, Na+, C2H3O21-, H2O Major species: HC2H3O2 (aq)  H+ (aq) + C2H3O21-(aq) I 0.50 M 0 0.50 M C -X +X +X E 0.50 – X X 0.50 + X Assume X’s = 0, and solve Notice when the [acid]=[base], then pH = pKa = 1.8 x 10 -5

  4. Buffers made easy Henderson-Hasselbalch equation • Used for buffer solutions Same problem = 4.74

  5. Now with the addition 0.010 mol NaOH to 1 L of 0.50 M HC2H3O2/NaC2H3O2 ms: HC2H3O2, Na+, OH-, C2H3O21-, H2O Mixing an acid, HC2H3O2, with OH- HC2H3O2 (aq) + OH- C2H3O21- (aq) + H2O (l) I.C.E. must be in MOLES I. 0.50 mol 0.010 mol 0.50 mol C. -0.010 -0.010 +0.010 E. 0.49 0 0.51 = 4.74 + log(.51/.49) = 4.76

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