Top Banner
Chapter 11 – Chemical Equilibrium
23

Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

Dec 16, 2015

Download

Documents

Bathsheba Marsh
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

Chapter 11 – Chemical Equilibrium

Page 2: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

Homework Assignment, Ch 8 (buffers)

Problems 5,6,9,11,12,13,18,19,20

Due Fri, Nov 1

Page 3: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

Exam 2 on Wed, Nov 6

Covers Chapters 5, 6, 7, 8 plus

Gravimetric and Volumetric Chloride

Determinations

Page 4: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

Homework Assignment, Ch 12 (EDTA)

Problems 1,8,10,12,17,23

Due Fri, Nov 8

Page 5: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

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 precipitation.

Page 6: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.
Page 7: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

Chapter 11 – Chemical Equilibrium

This effect of the ionic environment within the solution is known as the ionic strength and may be represented as (your author) or I (other authors).

= I = ½ (c1z12 + c2z2

2 +…) = ½ cizi

2

The sum of terms includes all of the ions in solution. An example of this calculation is shown in Problem 1

Page 8: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

What is the ionic strength of a solution that is 0.0100 M in KNO3 and 0.0100 M Na2SO4?

= ½ cizi2

= ½ {0.01(+1)2 + 0.01(-1)2 + 0.02(+1)2 + 0.01(-2)2 }

= ½ {0.08} = 0.04M

Page 9: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

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

Page 10: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

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 < == > cC + dD

K = (ac)c(aD)d / (aA)a(aB)b

or K = ([C] C)c([D] D)d / ([A] A)a([B] B)b

Page 11: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

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))} (Eq11-5)

where is the size of the ion in pm (picometers). Examples of for the F- and I- ions are shown in the next slide.

Page 12: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.
Page 13: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

Activity Coefficients

A more general form of the extended Debye-Huckel equation is

log 10 = -0.509z2 / (1 + ) *

Although this equation is less exact, one does not need all of the parameters of the previous form.

* From Hargis, “Analytical Chemistry: Principles and Techniques, p 19, Prentice-Hall, 1988.

Page 14: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

Activity Coefficients

The effects of ionic strength on activity for various charges of ions are shown in the next slide.

Page 15: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

Figure 11-4 as a function of for different values of z

Page 16: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

What is the activity coefficent of Mg+2 in a 3.3 mM solution of Mg(NO3)2, using both the author’s equation 11-5 and the given simplified forms?

In both, we first need the ionic strength :

= ½ {(3.3mM)(+2)2 + (6.6mM)(-1)2}

= ½ {13.2 + 6.6} = ½{19.8} = 9.9 mM

= 0.0099 M

Page 17: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

What is the activity coefficent of Mg+2 in a 3.3 mM solution of Mg(NO3)2, using both the author’s equation 11-5 and the given simplified forms?

Using the author’s eqn 11-5,

log = (-0.51)(+2)2(0.0099)1/2 / {1 +(800)(0.0099)1/2/305}

log = - 0.203/1.261 = -0.161

= 10 -0.161 = 0.690

Page 18: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

What is the activity coefficent of Mg+2 in a 3.3 mM solution of Mg(NO3)2, using both the author’s equation 11-5 and the given simplified forms?

Using the simplified equationlog = (-0.509)(+2)2(0.0099)1/2 / {1 +(0.0099)1/2}

log = - 0.203/1.099 = -0.185

= 10-0.185 = 0.654

Page 19: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

What is the activity coefficent of Mg+2 in a 3.3 mM solution of Mg(NO3)2, using both the author’s equation 11-5 and the given simplified forms?

In summary, the more exact equation 11-5 gives = 0.690 while the the simplified equation gives a value of = 0.654. In most cases the difference between the 2 values would not be important.

Page 20: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

Activity Coefficients

- Ignore activity coefficients for nonionic compounds.

Page 21: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

Activity Coefficients

Whenever the ionic strength is high ( > 1 M) the activity coefficient values we could calculate are not very meaningful because of the lower concentration of the solvent and the concentrations of the solutes increase.

Page 22: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

Figure 11-5 of H+ in 0.010 M HClO4 as a function of [NaClO4]

Page 23: Chapter 11 – Chemical Equilibrium. Homework Assignment, Ch 8 (buffers) Problems 5,6,9,11,12,13,18,19,20 Due Fri, Nov 1.

pH

The response of the glass electrode for measurement of pH is dependent on the ionic strength of the solution. The technical definition of pH is

pH = - log 10 (aH+) = log 10 H+ [H+] Also, be aware that most glass electrodes show significant errors at pH > 12 because of the high concentrations of the counter ions such as Na+ and K+.