Now Let Us Do the Setup Part of One More Complex Equilibrium Problem

Lecture 6 October 23, 2002

Today I want to do the setup part of one more complex equilibrium problem. Then I’ll talk a bit about the exam and we will work some review/homework problems.

Calculate the solubility of silver sulfide in pure water. [How do we know what the ions are? Table on p. 154.] XXXXX

Our old method Let x = [S2-]

+ 2- Ag2S(s) < == > 2 Ag + S

Before solid 0 0 After solid 2x x

-51 + 2 2- 2 3 3 -51 -17 Ksp = 8 x 10 = [Ag ] [S ] = (2x) x = 4x  x = 2 x 10  x = 1.3 x 10 M

Step 1. Pertinent Equilibria

+ 2- Ag2S(s) < == > 2 Ag + S 2- - - S + H2O < == > HS + OH - - HS + H2O < == > H2S + OH + - H2O < == > H + OH

Step 2. Definition of Unknown

+ 2- - solubility = ½ [Ag ] = [S ] + [HS ] + [H2S]

Step 3. Equilibrium-Constant Expressions

+ 2 2- -51 [Ag ] [S ] = Ksp = 8 x 10 (1)

[HS - ][OH -] = 0.769 (2) [S2-]

- -7 [H2 S][OH ] = 1.04 x 10 (3) [HS-] and [H+] [OH-] = 1.00 x 10-14 (4)

Step 4. Mass-Balance Expression same as Step 2 2

+ 2- - ½ [Ag ] = [S ] + [HS ] + [H2S] (5)

Step 5. Charge-Balance Expression

[Ag+] + [H+] = 2[S2-] + [HS-] + [OH-] (6) XXXXX

Step 6. Comparison of Equations and Unknowns We have six unknowns and six equations. Thus, an exact solution is feasible.

Step 7. Approximations

The solubility product for Ag2S is very small; therefore, it is probable that there is little change in hydroxide ion concentration as the precipitate dissolves. As a consequence, we can assume tentatively that

[OH-]  [H+] = 1.0 x 10-7 this means we can eliminate equation 4

Step 8. Solution of Equations

We now proceed by substitution of 1.0 x 10-7 for [OH-] in Equations (2) and (3).

[HS -] = 0.769 = 0.769 = 7.69 x 106  [HS-] = 7.69 x 106 [S2-] (2) [S2-] [OH-] 1.0 x 10-7

-7 - [H2S] = 1.04 x 10 = 1.04  [H2S] = 1.04 [HS ] (3) [HS-] 1.0 x 10-7

Substituting (2) to (3) yields

6 2- 6 2- [H2S] = 1.04 x 7.69 x 10 [S ] = 8.00 x 10 [S ]

Now we go back to Equation (5) and get

½ [Ag+] = [S2-] + (7.69 x 106)[S2-] + (8.00 x 106)[S2-] = 1.57 x 107[S2-]

[S2-] = [Ag + ] = 3.19 x 10-8 [Ag+] 2 x 1.57 x 107

Finally we use Equation (1) to get

[Ag+]2 (3.19 x 10-8[Ag+]) = 8 x 10-51 XXXX 3

[Ag+] = (8 x 10-51/3.19 x 10-8)1/3 = (2.51 x 10-43)1/3 = 6.3 x 10-15 solubility = ½ [Ag+] = 6.3 x 10 - 15 = 3 x 10-15 mol/L 2 step 9. Check of Assumptions

It is clear that the amount of silver sulfide going into solution is very small. This means that the total of all the sulfide species and silver ions remain at least 7 orders of magnitude below that of the hydrogen and hydroxide ion concentrations and therefore our assumption was valid.

Thus taking into account the interaction of the sulfide ion with water shows that the solubility of silver sulfide is 300x higher than we calculated before.

At least everything dilute that we don’t have to worry about activity effects.

Now, let me talk a bit about the exam.

Don't Panic!

An exam is a learning device. If you get something wrong, go back over it and learn from your mistakes. Use your study group to learn.

First, it will be in class, open book but not notes, homework, workshops, etc. Just your book. So you need to know where things are in your book.

Second you will need a functioning calculator. I strongly recommend that you have and know how to use a calculator that can calculate standard deviations. So it should have at least some basic statistics functions.

Most scientific calculators in the $10-20 range have such functions. So if you do not have one, buy one or borrow one. Next, learn how to use them before Monday! It can save you lots of time.

The maximum number of experimental values will be 5, so you can calculate standard deviations by hand if you don't have such a calculator, but it would be foolish to do so.

I give lots of partial credit so show all your work. If you make a mistake just cross-out what you don't want me to read and write a revised answer. If your revised answer is wrong I will go back to your first attempt to see if you have a better approach the first time. I am more interested in what you know and how you approach problems as getting right answers. 4

I don't want time to be a big problem, as I know that some of you panic when taking tests. I wrote the exam to take about an hour to an hour and a half. But I will be here by 8 AM and you can start then. You must finish by 10:30 when Jim starts his lecture so that should help those that feel time weighting on them.

In doing an exam of this type, you should first read through the exam and determine which questions are easy and which are harder for you.

Do the easy ones first.

Some questions have parts that are easy and parts that are hard. Do the easy ones first.

If you are clueless in Olympia. I offer you the following option. Tell me what you do know about a problem and how to approach answering the question. Do this only as a last resort, but do at least tell me what you do know.

If all else fails, define as many terms in the problem as possible. I do not expect to see any question left blank. A blank answer means that you don't know the means of the words "titration", milliliter, gram, nitrate, percent composition, etc.

The exam has 5 problems as of now – one equilibrium problem, two solubility problems, one uncertainty problem and one statistics problem.

I have given you most all of the formula weights you will need or references to pages in the book for some items. Others you will have to know how to find such as in the periodic table at the front of the book. So you must have a book.

Exam I Study Topics

*SI units *Chemical Concentrations – *molarity, *% composition, molality *Chemical Stoichiometry – formula mass, balancing chemical equations *Significant Figures *Logs, Antilogs and Exponents *Types of Uncertainties – systematic, random, precision, accuracy, absolute and relative uncertainty *Propagation of Uncertainty *Mean and Standard Deviation *Confidence Intervals Statistical treatment of data – t Test, Q test 5

*The Equilibrium Constant – manipulating constants of related reactions Equilibrium and Thermodynamics *Le Chatelier’s Principle *Solubility Product *The Common Ion Effect *Approximations Complex ion formation Acids and Bases *The p function Titrations Activity *Ionic Strength Systematic Treatment of Equilibrium

Workshop homework solutions

1) We start with a 0.8040-g sample of iron ore. The ore sample is dissolved and the iron 2+ converted to the Iron(II) (Fe ) ion. The solution is titrated with a 0.02242 M KMnO4 solution. The initial buret reading is 0.30 mL and the reading at the end point is 47.52 mL. The balanced equation for the reaction is

- 2+ + 2+ 3+ MnO4 + 5 Fe + 8 H == > Mn + Fe + 4 H2O a) What is the percent iron in the ore? b) If we have the following uncertainty in the measured quantities: mass  0.0001 g volume  0.02 mL molarity  0.00003 M. What is the uncertainty in the percent iron in the ore?

The first part of the problem is a stochiometry problem. Let’s look at the volume first.

47.52 – 0.30 mL = 47.22 mL

The uncertainty in an added or subtracted reading is found from the absolute uncertainties: . . Uncertainty in answer = √( 0.02)2 + ( 0.02)2 =  0.0283 mL

- - 47.22 mL MnO4 x 0.02242 mmol MnO4 x 5 mmol iron x 55.845 mg iron = - - mL MnO4 1 mmol MnO4 mmol iron

295.608 mg Fe in sample or 0.295608 g Fe 6

0.295608 g iron x 100% = 36.77 % iron in the ore sample 0.8040 g sample

The uncertainty in multiplication and/or division is found from percent relative errors: Volume 0.0283/47.22 x 100% =  0.06% Concentration 0.00003/0.02242 x 100% =  0.134% Mass 0.0001/0.8040 x 100% =  0.012% Atomic Weight 0.002/55.845 x 100% =  0.004% . . Uncertainty in answer = √ ( 0.06)2 + (0.134)2 + (0.012)2 + (0.004)2 =  0.1474%

= 0.001474

36.77 x  0.1474% = 36.77  0.05 % Fe in ore

There is a very important utility to this exercise. Which measurement contributes the most uncertainty to the answer? If you want to decrease the uncertainty, which measurement needs to be improved?

Got to here

o 2) The pKsp at 25 C and zero ionic strength of lanthanum iodate, La(IO3)3 is 10.99.

3+ - La(IO3)3 < == > La + 3 IO3

Assume that La(III) and iodate are the only ions formed when the salt dissolves. What is the solubility of lanthanum iodate (FM 663.63) under these conditions?

-10.99 -11 Ksp = 10 = 1.02 x 10

The molar solubility of lanthanum iodate equals the concentration of lanthanum ion since one lanthanum ion results for each formula unit dissolved.

3+ - La(IO3)3 < == > La + 3 IO3 Before solid 0 0 After solid x 3x

3+ - 3 -11 3 4 4 -11 Ksp = [La ][IO3 ] = 1.02 x 10 = x(3x) = 27x  x = (1.02 x 10 )/27

-4 x = 7.85 x 10 M x 663.63 g/mol = 0.52 g La(IO3)3 /L

Homework These two problems and the following from Chapter 9: 2, 5, 8