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Chapter 11 Chemical Equilibrium ... For the reaction aA bB == cC dD. K = (ac)c(aD)d / (aA)a(aB)b ... Figure 11-4. as a function of for different ionic ... – PowerPoint PPT presentation

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Title: Announcements


1
Announcements
  • Homework Assignments
  • Volumetric Chloride Lab due Monday, Oct 20

2
Chapter 11 Chemical Equilibrium
  • Although we have already used concepts of
    equilibrium in some of the previous chapters, we
    want to take a more detailed look at this
    important topic.

3
Chapter 11 Chemical Equilibrium
  • Chemical equilibrium describes the distribution
    of the products and reactants
  • in a reaction mixture. Frequently these
    relationships are more complicated than first
    meets the eye. For example, the equilibria
    (plural) for the solution of slightly soluble
  • lead(II) iodide in water is shown in the next
    slide.

4
The percent values give the approximate
contribution of each species to the whole.
5
Chapter 11 Chemical Equilibrium
  • Although the model of chemical equilibrium
    predicts that inert, that is, non-reactive
    salts should not affect the solubility, the
    actual effect is shown in the next slide, where
    the addition of KNO3 has caused an increase in
    the solubility of PbI2. In general the addition
    of such an inert salt causes an increase the
    solubility.

6
Increased solubility of PbI2. in KNO3
7
Chapter 11 Chemical Equilibrium
  • Our understanding of the phenomena shown on the
    proceeding slide is that the inert salt increases
    the ionic atmosphere (environment), allowing each
    cation or anion to be surrounded by species of
    the opposite charge, but farther separated from
    the counter ion which caused its original
    chemical reaction.

8
Ionic Environment around Cations or Anions.
9
Chapter 11 Chemical Equilibrium
  • The effect of the ionic environment within the
    solution is known as the ionic strength and may
    be represented as ? or I.
  • ? I ½ (c1z12 c2z22 ) ½ ? cizi2
  • The sum of terms includes all of the ions in
    solution. An example of this calculation is
    shown in Problem 1

10
What is the ionic strength of a solution that is
0.0100 M in KNO3 and 0.0100 M Na2SO4?
  • All of the ions present contribute to the ionic
  • environment.
  • I ? ½ ? cizi2
  • ? ½ 0.01(1)2 0.02(1)2 0.01(-1)2
    0.01(-2)2
  • K Na
    NO3? SO4 ?2
  • ? ½ 0.08 0.04M

11
Activity Coefficients
  • The actual or effective concentration of an
    ionic species in solution is known as the
    activity your author uses the symbol A (more
    commonly used is simply a lower case a)
  • I will use the later symbol, so that his
  • Equation 11-2 is written as
  • aC C ?C

12
Activity Coefficients
  • The exact equilibrium constant K is then
    expressed in terms of the activities of the
    species involved instead of the more commonly
    concentrations.
  • For the reaction aA bB lt gt cC dD
  • K (ac)c(aD)d / (aA)a(aB)b
  • or K (C ?C)c(D ?D)d / (A ?C)a(B ?C)b

13
Activity Coefficients
  • The individual values for the activity
    coefficients ? of each of the species is a
    function of the ionic strength as shown by the
    extended Debye-Huckel equation
  • __
    __
  • log 10 ? -0.51z2? ? / 1 (??? (305))
  • where ? is the size of the ion in pm
    (picometers). Examples of ? for the F- and I-
    ions are shown in the next slide.

14
? Values for the F- and I- ions where ?
represents the hydrated radius of the ions.
15
Figure 11-4 ? as a function of ? for different
ionic charges, values, z.
16
The activity coefficient ? for the H ion as a
function of the ionic strength.
17
The Refined definition of pH
  • Because of the effect ionic strength may have on
    the activity of an ion in an environment of other
    ions, the definition of pH is refined as
  • pH - log10 aH - log10 H ?H

18
Charge balance in a solution that contains 0.0250
M KH2PO4 and 0.0300 M KOH. The sum of positive
charges sum of negative charges.
19
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