# Chapter 17: Additional Aspects of Aqueous Equilibria

## Presentation on theme: "Chapter 17: Additional Aspects of Aqueous Equilibria"— Presentation transcript:

Chapter 17: Additional Aspects of Aqueous Equilibria
Section 1: The Common Ion Effect

Objectives When you complete this presentation, you will be able to
describe the common ion effect.

Introduction Water is the most common and most important solvent on Earth. We will be looking in some detail at the application of equilibrium theory and practice to aqueous solutions. Additional acid-base equilibria Buffers and acid-base titrations Solubility of compounds Formation of complex ions

The Common-Ion Effect We know that sodium salts are strong electrolytes and dissociate completely in aqueous solution. NaA(aq) → Na+(aq) + A−(aq) We also know that certain acids are weak electrolytes and dissociate partially in solution. HA(aq) ⇄ H+(aq) + A−(aq)

The Common-Ion Effect If we start with a solution of acetic acid, we will set up the equilibrium for the weak acid. HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq) If we add sodium acetate to the solution, the additional acetate will drive the equilibrium to the left, decreasing the equilibrium [H+]. The presence of the added acetate ion causes the acetic acid to ionize less than it normally would.

The Common-Ion Effect Whenever a weak electrolyte and a strong electrolyte contain a common ion, the weak electrolyte ionizes less than it would if it were alone in the system. This is called the common-ion effect. We can calculate equilibrium concentrations of systems with common ions.

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? Plan: Identify the major species in solution. Identify the major sources that affect the concentration of H+. Build “i-c-e” table. Use K expression to calculate [H+].

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? Major species: HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq)

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? Major sources that affect the concentration of H+: HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq)

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? “i-c-e” table: HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq) initial 0.30 M 0.00 M change -x +x equilibrium (0.30 – x) M x M ( x) M

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? K expression: Ka = 1.8 × 10−5 = [H+][CH3COO−] [HCH3COO]

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? K expression: Ka = 1.8 × 10−5 = x( x) 0.30 − x

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? K expression: Ka = 1.8 × 10−5 = x( x) 0.30 − x assume x<<0.30

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? K expression: Ka = 1.8 × 10−5 = x(0.30) 0.30 assume x<<0.30

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? K expression: Ka = 1.8 × 10−5 = x(0.30) 0.30

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? K expression: Ka = 1.8 × 10−5 = = x x(0.30) 0.30

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? K expression: Ka = 1.8 × 10−5 = = x = [H+] x(0.30) 0.30

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? calculate pH: pH = −log[H+]

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? calculate pH: pH = −log[H+] = −log(1.8 × 10−5)

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? calculate pH: pH = −log[H+] = −log(1.8 × 10−5) = 4.74

The Common-Ion Effect Sample Exercise 17.1 (pg. 720)
What is the pH of a solution made by adding mol of acetic acid and 0.30 mol of sodium acetate to enough water to make a 1.0 L solution? calculate pH: pH = −log[H+] = −log(1.8 × 10−5) = 4.74

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Plan: Identify the major species in solution. Identify the major sources that affect the concentration of H+ & F−. Build “i-c-e” table. Use K expression to calculate [H+] & [F−].

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Major species: HF(aq) ⇄ H+(aq) + F−(aq)

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Major sources that affect [H+]: HF(aq) ⇄ H+(aq) + F−(aq)

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. “i-c-e” table: HF(aq) ⇄ H+(aq) + F−(aq) initial 0.20 M 0.10 M 0.00 M change -x +x equilibrium (0.20 – x) M ( x) M x M

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. K expression: Ka = 6.8 × 10−4 = [H+][F−] [HF]

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. K expression: Ka = 6.8× 10−4 = (0.10 +x)x 0.20 − x

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. K expression: Ka = 6.8× 10−4 = (0.10 +x)x 0.20 − x assume x<<0.30

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. K expression: Ka = 6.8× 10−4 = (0.10)x 0.20 assume x<<0.30

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. K expression: Ka = 6.8× 10−4 = x = (6.8× 10−4) (0.10)x 0.20 0.20 0.10

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. K expression: Ka = 6.8× 10−4 = x = (6.8× 10−4) = [F−] (0.10)x 0.20 0.20 0.10

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. K expression: Ka = 6.8× 10−4 = x = (6.8× 10−4) = [F−] = 1.4 × 10−3 M (0.10)x 0.20 0.20 0.10

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. K expression: Ka = 6.8× 10−4 = x = (6.8× 10−4) = [F−] = 1.4 × 10−3 M (0.10)x 0.20 0.20 0.10

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Find pH: [H+] = 0.10 M − 1.4 × 10−3

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Find pH: [H+] = 0.10 M − 1.4 × 10−3 ≈ 0.10 M

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Find pH: [H+] = 0.10 M − 1.4 × 10−3 ≈ 0.10 M pH = −log[H+]

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Find pH: [H+] = 0.10 M − 1.4 × 10−3 ≈ 0.10 M pH = −log[H+] = −log(0.10)

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Find pH: [H+] = 0.10 M − 1.4 × 10−3 ≈ 0.10 M pH = −log[H+] = −log(0.10) = 1.00

The Common-Ion Effect Sample Exercise 17.2 (pg. 722)
Calculate [F−] and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. Find pH: [H+] = 0.10 M − 1.4 × 10−3 ≈ 0.10 M pH = −log[H+] = −log(0.10) = 1.00

The Common-Ion Effect Sample Exercises 17.1 and 17.2 both involve weak acids. We can also use the same techniques with weak bases. For example, adding NH4Cl to an aqueous solution of NH3 will cause the NH3 to dissociate less and lower the pH.

The Common-Ion Effect The techniques of Sample Exercises 17.1 and may be used to solve homework problems and