Practice: Calculate the pH of a solution formed by mixing 200 mL of a 0.400 M C_{2}H_{5}NH_{2} solution with 350 mL of a 0.450 M C_{2}H_{5}NH_{3}^{+ }solution. (K_{b} of C_{2}H_{5}NH_{2 }is 5.6 x 10 ^{-4}).

Concept #1: The Henderson – Hasselbalch Equation.

Example #1: What is the pH of a solution consisting of 2.75 M sodium phenolate (C_{6}H_{5}ONa) and 3.0 M phenol (C_{6}H_{5}OH). The K_{a} of phenol is 1.0 x 10^{-}^{10}.

Practice: Calculate the pH of a solution formed by mixing 200 mL of a 0.400 M C_{2}H_{5}NH_{2} solution with 350 mL of a 0.450 M C_{2}H_{5}NH_{3}^{+ }solution. (K_{b} of C_{2}H_{5}NH_{2 }is 5.6 x 10 ^{-4}).

Example #2: What is the buffer component concentration ratio, [Pr ^{- }] / [HPr] , of a buffer that has a pH of 5.11. (The K_{a} of HPr is 1.30 x 10^{-5}).

Example #3: Over what pH range will an oxalic acid (H_{2}C_{2}O_{4}) / sodium oxalate (NaHC_{2}O_{4}) solution work most effectively? The acid dissociation constant of oxalic acid is 6.0 x 10^{-}^{2}.

a) 0.22 – 2.22 b) 1.00 – 3.00 c) 0.22 – 1.22 d) 2.0 – 4.0

Practice: Determine how many grams of sodium acetate, NaCH_{3}CO_{2} (MW: 82.05 g/mol), you would mix into enough 0.065 M acetic acid CH_{3}CO_{2}H (MW: 60.05 g/mol) to prepare 3.2 L of a buffer with a pH of 4.58. The K_{a} is 1.8 x 10** ^{-5}**.

Example #4: Which weak acid-conjugate base combination would be ideal to form a buffer with a pH of 4.74.

a) Cyanic acid and Potassium cynate (K_{a} = 4.9 x 10^{-10})

b) Benzoic acid and Lithium benzoate (K_{a} = 6.3 x 10^{-5})

c) Acetic acid and Sodium acetate (K_{a} = 1.7 x 10^{-5})

d) Ammonium chloride and Ammonia (K_{a} = 5.56 x 10^{-10})

e) Formic acid and Cesium formate (K_{a} = 1.7 x 10^{-4})

Practice: A buffer solution is made by combining a weak acid with its conjugate salt. What will happen to the pH if the solution is diluted to one-fourth of its original concentration?