Chapter 17 Additional Aspects of Aqueous Equilibria

# Chapter 17 Additional Aspects of Aqueous Equilibria

## Chapter 17 Additional Aspects of Aqueous Equilibria

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1. Chapter 17Additional Aspects of Aqueous Equilibria

2. Buffers • Buffers are solutions of a weak conjugate acid–base pair. • They are particularly resistant to pH changes, even when small additions of strong acid or base is added.

3. [base] [acid] pH = pKa + log • Henderson–Hasselbalch equation Acidic buffer: pH < 7 (acid stronger than base) Example: Acetic acid (1M, ka = 1.8x10-5) / acetate anion (1M, ka = 5.6x10-10) pH = pKa = 4.72

4. [base] [acid] pH = pKa + log • Henderson–Hasselbalch equation Basic buffer: pH > 7 (base stronger than acid) Example: NH4+ (1M, ka = 5.6x10-10) / NH3 (1M, ka = 1.8x10-5) pH = pKa = 9.25

5. [base] [acid] pH = pKa + log (0.10) (0.12) pH = log (1.4  104) + log Henderson–Hasselbalch Equation What is the pH of a buffer that is 0.12 M in lactic acid, CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka for lactic acid is 1.4  104. pH = 3.85 + (0.08) pH = 3.77

6. pH Range • The pH range is the range of pH values over which a buffer system works effectively. • It is best to choose an acid with a pKa close to the desired pH. • The larger the concentrations of the components of a buffer, the more effective it is at resisting change in pH. • (The number of moles of weak acid and base used to prepare the buffer must be considerably greater than the number of moles os acid or base that may later be added to the buffer)

7. Calculating pH Changes in Buffers A buffer is made by adding 0.300 mol HC2H3O2 (pKa = 1.8 x 10-5) and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. a) What is the pH of this buffer? b) Calculate the pH of this solution after 0.020 mol of NaOH is added, c) after 0.020 mol of HCl is added a) pH = pKa = log (1.8  105) = 4.74

8. (0.320) (0.280) pH = 4.74 + log Calculating pH Changes in Buffers b) The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC2H3O2(aq) + OH(aq)  C2H3O2(aq) + H2O(l) Before 0.30 0.0 0.30 Add 0.020 After 0.28 0.32 Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + 0.06 = 4.80 (very small increase in basesity), however, Without a buffer NaOH dissociates completely in solution pOH = -log(0.020) = 1.7, pH = 12.3

9. (0.280) (0.320) pH = 4.74 + log Calculating pH Changes in Buffers c) The 0.020 mol HCl will react with 0.020 mol of the acetate anion: C2H3O2(aq) + HCl(aq)  HC2H3O2(aq) + Cl-(l) Before 0.30 0.0 0.30 Add 0.020 After 0.28 0.32 Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 - 0.06 = 4.68 (very small increase in acidity), however, Without a buffer HCl dissociates completely in solution pH = -log(0.020) = 1.7