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## PowerPoint Slideshow about 'Chapter 17: Additional Aspects of Aqueous Equilibria' - keran

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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 0.30 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 0.30 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 0.30 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)
- “i-c-e” table:
- HCH3COO(aq) ⇄ H+(aq) + CH3COO−(aq)

The Common-Ion Effect

- Sample Exercise 17.1 (pg. 720)
- K expression:

Ka = 1.8 × 10−5 =

[H+][CH3COO−]

[HCH3COO]

The Common-Ion Effect

- Sample Exercise 17.1 (pg. 720)
- K expression:

Ka = 1.8 × 10−5 =

x(0.30 + x)

0.30 − x

The Common-Ion Effect

- Sample Exercise 17.1 (pg. 720)
- K expression:

Ka = 1.8 × 10−5 =

x(0.30 + x)

0.30 − x

assume x<<0.30

The Common-Ion Effect

- Sample Exercise 17.1 (pg. 720)
- 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)
- K expression:

Ka = 1.8 × 10−5 =

x(0.30)

0.30

The Common-Ion Effect

- Sample Exercise 17.1 (pg. 720)
- K expression:

Ka = 1.8 × 10−5 = = x

x(0.30)

0.30

The Common-Ion Effect

- Sample Exercise 17.1 (pg. 720)
- K expression:

Ka = 1.8 × 10−5 = = x = [H+]

x(0.30)

0.30

The Common-Ion Effect

- Sample Exercise 17.1 (pg. 720)
- calculate pH:

pH = −log[H+]

The Common-Ion Effect

- Sample Exercise 17.1 (pg. 720)
- calculate pH:

pH = −log[H+] = −log(1.8 × 10−5)

The Common-Ion Effect

- Sample Exercise 17.1 (pg. 720)
- calculate pH:

pH = −log[H+] = −log(1.8 × 10−5) = 4.74

The Common-Ion Effect

- Sample Exercise 17.1 (pg. 720)
- 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)

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 17.2 may be used to solve homework problems 17.15 and 17.17.

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