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AP Chemistry Unit 11 – Additional Equilibrium Topics Lesson 1 – The Common Ion Effect Book Section: 17.1.

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Presentation on theme: "AP Chemistry Unit 11 – Additional Equilibrium Topics Lesson 1 – The Common Ion Effect Book Section: 17.1."— Presentation transcript:

1 AP Chemistry Unit 11 – Additional Equilibrium Topics Lesson 1 – The Common Ion Effect Book Section: 17.1

2 Unit 11 – Additional Equilibrium Topics Lesson 1 – The Common Ion Effect (17.1) Lesson 2 – Buffers (17.2) Lesson 3 – Acid/Base Titrations (17.3) Lesson 4 – Solubility Equilibrium (17.4- 17.5) Lesson 5 – K sp (17.4-17.5) Lesson 6 – Precipitation (17.6)

3 “Common Ions” When we dissolve acetic acid in water, the following equilibrium is established:  CH 3 COOH  CH 3 COO - + H + If sodium acetate were dissolved in solution, which way would this equilibrium shift?

4 “Common Ions” When we dissolve acetic acid in water, the following equilibrium is established:  CH 3 COOH  CH 3 COO - + H + If sodium acetate were dissolved in solution, which way would this equilibrium shift?  Adding acetate ions from a strong electrolyte would shift this equilibrium left (Le Chatelier’s principle).

5 Common Ion Effect Whenever a weak electrolyte (acetic acid) and a strong electrolyte (sodium acetate) share a common ion, the weak electrolyte ionizes less than it would if it were alone. (Le Chatelier’s) This is called the common-ion effect.

6 Steps for Common-Ion Problems 1. Consider which solutes are strong electrolyte and weak electrolytes. 2. Identify the important equilibrium (weak) that is the source of H + and therefore determines pH. 3. Create an ICE chart using the equilibrium and strong electrolyte concentrations. 4. Use the equilibrium constant expression to calculuate [H + ] and pH.

7 Sample Problem What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution? K a for acetic acid = 1.8 x 10 -5

8 Sample Problem Calculate the fluoride concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl K a for HF = 6.8 x 10 -4.

9 Homework: 17.14, 17.16, 17.18 Thursday: Buffers (17.2) Friday: No Class Problem Set 9 – 3/21 (Mon) Acid/Base Exam – 3/22 (Tues) Titration 3 – 3/24 (Thurs) It’s crunch time! 19 more class days of AP Chemistry!


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