Ap Chem Chapter 17 Outline

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AP Chemistry

Chapter 17 Additional Aspects of Aqueous Equilibria

Chapter 17. Additional Aspects of Equilibrium
Common Student Misconceptions






Students often believe that the pH at the equivalence point for any titration is 7.00. In terms of problem-solving skills, this is probably the most difficult chapter for most students. Students tend to find buffers particularly difficult to understand. Students often forget to consider volume changes that occur when two solutions are mixed (this will have an effect on the concentration of the species present).

Students tend to confuse Ksp and solubility.

17.1 The Common Ion Effect


The dissociation of a weak electrolyte is decreased by the addition of a strong electrolyte that has an ion in common with the weak electrolyte.



For example, consider the ionization of a weak acid, acetic acid. HC2H3O2(aq) ⇋ H+(aq) + C2H3O2–(aq)




This causes a reduction in the [H+] and a decrease in the percent ionization of the acetic acid.



By adding sodium acetate, we have disturbed the acetic acid equilibrium.





If we add additional C2H3O2– ions by the addition of a strong electrolyte, (e.g., NaC2H3O2) the equilibrium is shifted to the left.

In effect, we have added a product of this equilibrium (i.e., the acetate ion). • This phenomenon is called the common-ion effect.

Common ion equilibrium problems are solved following the same pattern as other equilibrium problems. • However, the initial concentration of the common ion (from the salt) must be considered.

Sample Exercise 17.1 (p. 662)
What is the pH of a solution made by adding 0.30 mol of acetic acid (HC2H3O2) and 0.30 mol of sodium acetate (NaC2H3O2) to enough water to make 1.0 L of solution?
(4.74)

-1-

AP Chemistry

Chapter 17 Additional Aspects of Aqueous Equilibria

Practice Exercise 17.1
Calculate the pH of a solution containing 0.085 M nitrous acid (HNO2, Ka = 4.5 x 10-4) and 0.10 M potassium nitrite (KNO2).
(3.42)

Sample Exercise 17.2 (p. 663)
Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl. ([F-] = 1.4 x 10-3 M; pH = 1.00)

-2-

AP Chemistry

Chapter 17 Additional Aspects of Aqueous Equilibria

Practice Exercise 17.2
Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid (HCHO2; Ka = 1.8 x 10-4) and 0.10 M in HNO3.
([CHO2-] = 9.0 x 10-5 M; pH = 1.00)

17. 2 Buffered Solutions


A buffered solution or buffer is a solution that resists a change in pH upon addition of small amounts of strong acid or strong base.

Composition and Action of Buffered Solutions


A buffer consists of a mixture of a weak acid (HX) and its conjugate base (X– ). HX(aq) ⇋ H+(aq) + X– (aq)



Thus a buffer contains both:
• An acidic species (to neutralize OH–) and
• A basic species (to neutralize H+).



When a small amount of OH– is added to the buffer, the OH– reacts with HX to produce X– and water. • But the [HX]/[ X–] ratio remains more or less constant, so the pH is not significantly changed.



When a small amount of H+ is added to the buffer, X– is consumed to produce HX. • Once again, the [HX]/[ X– ] ratio is more or less constant, so the pH does not change significantly. -3-

AP Chemistry

Chapter 17 Additional Aspects of Aqueous Equilibria

Buffer Capacity and pH


Buffer capacity is the amount of acid or base that can be neutralized by the buffer before there is a significant change in pH.



Buffer capacity depends on the concentrations of the components of the buffer. • The greater the concentrations of the conjugate acid-base pair, the greater the buffer capacity.



The pH of the buffer is related to Ka and to the relative concentrations of the acid and base.



We can derive an equation that shows the relationship between conjugate acid-base concentrations, pH and Ka.
By definition:



Ka =


[H...
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