16.2 Buffers: Solutions That Resist pH Change Casey Schulkind

Buffers Buffer: a solution that resists pH change by neutralizing added acid or base Buffer Characteristics: Resists pH change & can maintain a nearly constant pH When additional acid is added to a buffer, the conjugate base reacts with the acid, neutralizing it Contains significant amounts of both a weak acid and its conjugate base (or a weak base and its conjugate acid) The weak acid neutralizes added base The base neutralizes added acid

Buffer consisting of a weak base and its conjugate acid Buffer containing a weak acid (which can neutralize added base) and its conjugate base (which can neutralize added acid) Buffer consisting of a weak base and its conjugate acid

Practice Which solution is a buffer? A solution that is 0.100 M in HNO₂ and 0.100 M in HCl A solution that is 0.100 M in HNO₃ and 0.100 M in NaNO₃ A solution that is 0.100 M in HNO₂ and 0.100 M in NaCl A solution that is 0.100 M in HNO₂ and 0.100 M in NaNO₂ (d) A solution that is 0.100 M in HNO₂ and 0.100 M in NaNO₂

Calculating the pH of a Buffer Solution Common Ion Effect: the tendency of a common ion to decrease the solubility of an ionic compound or to decrease the ionization of a weak acid or weak base To find the pH of a buffer solution containing common ions: Work an equilibrium problem in which the initial concentrations include both the acid and its conjugate base

HC₂H₃O₂ (aq) + H₂O (l) ⇋ H₃O⁺ (aq) + C₂H₃O₂⁻ (aq) Practice Calculate the pH of a buffer solution that is 0.100 M in HC₂H₃O₂ and 0.100 M in NaC₂H₃O₂. Kₐ = 1.8 x 10⁻⁵ HC₂H₃O₂ (aq) + H₂O (l) ⇋ H₃O⁺ (aq) + C₂H₃O₂⁻ (aq) [HC₂H₃O₂] [H₃O⁺] [C₂H₃O₂⁻] I 0.100 ~0 C -x +x E 0.100 - x x 0.100 + x

(continued) X is Small: Kₐ = [H₃O⁺][C₂H₃O₂⁻] / [HC₂H₃O₂] 1.8 x 10⁻⁵ = (x)(0.100+x) / (0.100-x) 1.8 x 10⁻⁵ = x X is Small: ((1.8 x 10⁻⁵) / 0.100) x 100% = 0.018% pH= -log[H₃O⁺] pH = -log(1.8 x 10⁻⁵) pH = 4.74 [H₃O⁺] = x = 1.8 x 10⁻⁵

The Henderson-Hasselbalch Equation pH = pKₐ + log([base] / [acid]) Allows us to calculate the pH of a buffer solution from the initial concentrations of the buffer components *can be used only if x is small and if you are given the concentrations of both acid and base

Practice Calculate the pH of a buffer solution that is 0.050 M in benzoic acid (C₇H₆O₂) and 0.150 M in sodium benzoate (NaC7H5O2). Kₐ=6.5 x 10-5 pH = pKₐ + log([base] / [acid]) pH = -log(6.5 x 10-5) + log(0.150 / 0.050) pH = 4.187 + 0.477 pH = 4.66

Calculating pH Changes in a Buffer Solution Although a buffer resists pH change as acid or base is added to it, the pH does change by a small amount Calculating the pH change requires breaking up the problem into two parts: Stoichiometric calculation: to calculate how the addition changes the relative amounts of acid and conjugate base Instead of an ICE table, set up a “before addition, addition, after addition” table to track stoichiometric changes that occur during the neutralization of the added acid. Equilibrium calculation: calculate the pH based on the new amounts of acid and conjugate base Set up an ICE table, using the “after addition” values as initial concentrations

Practice A 1.0L buffer solution contains 0.100 mol HC₂H₃O₂ and 0.100 mol NaC₂H₃O₂. Kₐ = 1.8 x 10-5. Because the initial amounts of acid and conjugate base are equal, pH = pKₐ = -log(1.8 x 10-5) = 4.74. Calculate the new pH after adding 0.010 mol of solid NaOH to the buffer. Part 1: Stoichiometry Part 2: Equilibrium HC₂H₃O₂ + H2O⇌ H3O+ + C₂H₃O₂- OH- HC₂H₃O₂ C₂H₃O₂- Before addition ~0.00 mol 0.100 mol Addition 0.010 mol --- After Addition 0.090 mol 0.110 mol [HC₂H₃O₂] [H3O+] [C₂H₃O₂-] I 0.090 ~0 0.110 C -x +x E 0.090 - x x 0.110 + x

(continued) pH = -log(1.47 x 10-5) Kₐ = [H3O+][C₂H₃O₂-] / [HC₂H₃O₂] Kₐ = x(0.110+x) / 0.090-x 1.8 x 10-5 = 0.110x / 0.090 X = [H3O+] = 1.47 x 10-5 pH = -log(1.47 x 10-5) pH = 4.83 X is Small: (1.47 x 10-5 / 0.090) x 100% = 0.016%

Buffers Containing a Base and its Conjugate Acid When a buffer is composed of a base and its conjugate acid, the conjugate acid is the ion To calculate the pH of a solution using the Henderson-Hasselbalch equation, first calculate pKₐ for the conjugate acid of the weak base *Kₐ x Kb = Kw *pKₐ + pKb = 14