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NAME_________________________________________________________ PERIOD_________

ACTIVE LEARNING IN CHEMISTRY EDUCATION

"ALICE"

APPENDICES

A-1 ©1997, A.J. Girondi

NOTICE OF RIGHTS

All rights reserved. No part of this document may be reproduced or transmitted in any form by any means,electronic, mechanical, photocopying, or otherwise, without the prior written permission of the author.

Copies of this document may be made free of charge for use in public or nonprofit private educationalinstitutions provided that permission is obtained from the author . Please indicate the name and addressof the institution where use is anticipated.

© 1997 A.J. Girondi, Ph.D.505 Latshmere DriveHarrisburg, PA 17109

[email protected]

Website: www.geocities.com/Athens/Oracle/2041


APPENDIX A

SECTION A-1. Uncertainty in measurement

Measurement is the foundation of all experimental science. Accordingly, the degree of certaintyabout conclusions drawn from experiments depends upon the accuracy and precision of themeasurements made while conducting the experiments. The accuracy of a measurement indicates theamount by which it differs from a known, true value; the precision of a measurement is a measure of the"reproducibility" of the measurement. Let's use an analogy to make this more clear. Suppose weconsider two archers who are shooting arrows at the targets below. The small circles in the centers of thetargets are the bull's eyes, and the x's mark the points where the archers' arrows have struck the targets.Note that archer A shot arrows all over the place. Archer B shot arrows which landed very close together,but not near the bull's eye. Archer C shot arrows very close together too, and they also hit the desiredspot! Archer B is precise, but not accurate. Archer C is both precise and accurate. Archer A should sellhis bow before he hurts somebody!

xx

x

xx

xxxx x

x xxx x

Archer A Archer B Archer C

Figure A-1

Archer B is precise because he can reproduce his results. You can also think of precision as areflection of uncertainty. If precision is high, uncertainty is low. Since archer B is precise, the uncertaintyabout where his arrows will hit is small. Time after time, he gets the same result. (Even though he maynever hit the bull's eye.) Archer C always gets the desired result. Therefore, he is accurate. Since hisaccuracy is consistent, he is also precise. Archer A is not precise or accurate, poor guy.

The scientific investigator tries to be both accurate and precise. However, regardless of theexactness of the measurements, there is a degree of uncertainty. This uncertainty may be caused by theuse of inadequate instruments, by inexactness in the readings made by the investigator, or by acombination of these factors. It is customary in reporting scientific measurements to include all of thefigures known with certainty and one doubtful or estimated figure. For example, the metric ruler shownbelow in Figure A-2 can be read as 3.5 cm.

Since the 3 can be read directly from the ruler, it is known with certainty. The 0.5 cm reading is anestimate made by the observer and is only the estimate of the person reading the scale. Althoughestimated and therefore doubtful, the 0.5 cm is reported as part of the measurement . The digits knownwith certainty in any measurement and the first estimated digit are all part of the measurement and arecalled significant figures.

0 2 4 6 8 10

Figure A-2


It is important that a person does not report a figure such as 3.58 cm as having been read on theruler shown above. Since the 0.5 figure is only an estimate, the next figure (0.08) cannot even beestimated and, therefore, a measurement of 3.58 indicates more information than is really known.Similarly, a reported reading of 3 cm would convey too little information, because the reader would assumethat the 3 was an estimate, when in fact it is known with certainty.

A scientific measurement should include all of the information possible, but not more than cantruthfully be recorded. Generally speaking, most measurements can be read to one decimal place beyondthe smallest divisions on the scale being read. Thus, if an instrument is calibrated so that the closestmarked divisions are at intervals of 1 cm (the ones column), readings can usually be made to the tenthscolumn with the last digit being an estimate. Thus, we read 3.5 cm from the scale in Figure A-2.

There are two methods for expressing how much uncertainty there is in a measurement. In thefirst method, the uncertainty is specifically expressed. For example, the reading from the scale in FigureA-2 above was 3.5 cm. We are unsure of the 5. All we can say is that the real value is somewhere between3 and 4. Thus the "real" value could actually fall into a range which is 0.5 cm above or below the 3.5 cmreading. We would, therefore, say that the uncertainty in the measurement is ± 0.5 cm. Themeasurement would be expressed as 3.5 cm ± 0.5.

The second method of expressing uncertainty involves the use of the proper number ofsignificant figures when expressing the measurement. In this method, the amount of uncertainty is notexpressed. Instead, the column in which the uncertainty exists is implied. For example, when we write themeasurement as 3.5 cm, the reader will assume that the last significant figure (the 5) is an estimate. Thereader then understands that the uncertainty is in the tenths column. The reader understands that theamount of uncertainty is at least ± 0.1, but it could be more (such as ± 0.2 or ± 0.3, etc.). This method ofexpressing uncertainty is not as specific, but it is a lot more convenient.

In measurements that have been reported correctly, all nonzero numbers are always significantfigures. A zero may or may not be a significant figure. Why is that, and how can we tell whether a zero issignificant or not? First, we should try to establish if the zero in question is serving merely to determine orlocate a decimal point. When a zero serves as a determiner or indicator of a decimal point (that is, if itserves as a "place holder"), it is NOT considered to be a significant figure. For example, consider thefigures shown below. In Figure A-3, note that the arrow is located between 10,000 and 20,000. As weread the scale we are , therefore, sure that the first digit in the measurement should be a 1. This is not anestimate, and it is not, therefore, the final sig fig. Now, we estimate that the second digit might be a 2. Sowe read 12,000 as the location of the arrow. Since the 2 is an estimate, it is the final significant digit in themeasurement. We can read no more digits from the scale. However, in order to put the decimal in theproper location we need some place holders. Zeros are used for that purpose; but, these zeros will not beread or estimated from the scale. They are NOT significant. They serve merely as place holders. Look atFigure A-4. Note that we can be sure that the arrow is located between 10,000 and 10,001 on the scale.Thus, we are sure that the first digit should be a 1, and we are sure that the next four digits should bezeros. (We read them from the scale.) Since the arrow appears to be about half way between the twovalues, we estimate that the measurement should be read as 10,000.5. The 5, however, is an estimateand is, therefore, the final sig fig. The zeros in this case, are not merely place holders. They aresignificant!

10,000 20,000 30,000 40,000

Figure A-3


Figure A-4

10,000 10,001

The following 4 rules should be observed in determining whether a digit is a significant figure. You will beexpected to know these rules!

1. Nonzero digits are always significant.

2. "Leading zeros" (zeros which appear in the front portion of a number) are never significant. Forexample, the number 0.0039 has two "sig figs." The zeros are used to locate the decimal point. We oftenput a zero to the left of the decimal in numbers which have a value less than one. It is not significant.

3. "Trapped zeros" (zeros which appear between significant digits in a number) are always significant. Forexample, the numbers 0.0304 and 203 both have three sig figs., while the number 800006 has six sigfigs.

4. "Trailing zeros" (zeros at the end of a number) may or may not be significant. They are significant only ifthe decimal point is expressed. If the decimal is understood (not showing), the trailing zeros are notsignificant. Thus, the numbers 1900. and 16.00 both contain four sig figs since the decimal point isexpressed (showing) in both cases. However, in a number with an understood decimal point, the finalzeros are just used to locate the decimal point and are NOT significant. Thus, the number 16,000 has twosig figs (the 6 is uncertain). If we express the decimal, 16000., the final zero is the uncertain digit and thenumber now has five sig figs. (Sometimes when a number ends with several zeros, the last significant onehas a bar over it. For example, if the second zero in 16,000 had a bar over it, the number would thencontain 4 sig figs.)

When performing arithmetic calculations involving measured values, we must express our resultsso that they contain only the number of significant figures justified by the uncertainty of the originalmeasurements. Thus, it is frequently necessary to round off numbers so that a result does not appear tobe more certain than the original measurements.

The following rules should be carefully observed when rounding off a measurement:

a. When the digit dropped is less than 5, the preceding digit remains unchanged; for example, 8.3734when expressed to three sig figs becomes 8.37.

b. When the digit dropped is 5 or more, the preceding digit is increased by 1; for example, 3.6287expressed to three sig figs becomes 3.63.

When measurements are added or subtracted, the results of the calculations should be rounded off to thecolumn containing the leftmost uncertain digit. For example,

ADD SUBTRACT

28.6 cm 287.56 g 327.33 cm 76.4 g 5891.212 cm 211.16 g ---> (211.2 g)6247.142 cm ---> (6247.1 cm)

In the addition problem, since 28.6 contains the leftmost uncertain digit (the 6 in the tenths column),6247.142 is rounded off to 6247.1. In the subtraction problem, since the 76.4 contains the leftmost


uncertain digit (the 4 in the tenths column), 211.16 is rounded off to 211.2.

When measured values are multiplied or divided, count the number of sig figs in each measurement. Theone which has the least number of sig figs determines how many sig figs will appear in the answer.

Example: 28 cm X 4728 cm = 132,384 cm2 = 130,000 cm2 (rounded)

Since 28 contains two sig figs and 4728 contains 4 sig figs, the answer must be rounded to two sig figswhich would be 130,000. Since the zeros in 130,000 only indicate the position of the decimal point, theyare not sig figs. 130,000 cm2 contains only two sig figs.

Here are a couple of rules to remember:

1. When experimental quantities (measurements) are multiplied or divided, the result is rounded to thesame number of sig figs as the measurement which contains the least number of sig figs.

2. When experimental quantities (measurements) are added or subtracted, the uncertain digit in the resultmust be in the same column as the leftmost uncertain digit in the original measurements.

This sounds confusing so look at the addition problem illustrated on page A-5. The 6 is theuncertain digit in 28.6 and it is in the tenths column. The second 3 is the uncertain digit in 327.33 and it isin the hundredths column. The second 2 is uncertain in 5891.212 and it is in the thousandth column. Ofthe three uncertain digits, the leftmost one is the 6 in 28.6. So, the answer should be rounded to thetenths column and becomes 6247.1. Do you now see why the answer to the subtraction problem isrounded off to the tenths column?

When doing a problem which involves both multiplication (or division) AND addition (orsubtraction), you should round to the correct number of sig figs only when switching from multiplication (ordivision) to addition (or subtraction). Thus, you should not round off any intermediate results while you arejust multiplying and/or dividing or when you are just adding and/or subtracting.

For example, try to solve the problem below with your calculator.

1.113 cm X 4.3 cm X 8.11 cm

2.00 cm X 1.00 cm - 4.5 cm + 6.32 cm = ??? cm

The best way to handle this calculation is to do all the multiplication and division and then correctly roundthe result using the proper rule. Then, do all the addition and subtraction and round that result to thecorrect number of sig figs using the proper rule. Finally, the two calculations can be combined androunded to obtain the final result. The correct answer to the problem is 21 cm when rounded properly.

SECTION A-2. Problems

Problem 1. Underline any digits in the following measurements which are NOT significant.

a. 2.4421 cm h. 42.0040 m o. 7080.0940 mg

b. 200.41 m i. 3000 mm p. 0.8 mg

c. 3.00 L j. 5.300 g q. 674 L

d. 0.10004 cm3 k. 00.0050050 mL r. 4000200.080 m

e. 0.0020 m2 l. 240 kg s. 767003 cm

f. 00.0030030 m m. 23,000.010 cm t. 97600 g

g. 108,090 cm n. 0.0060 mL u. 8740. mg


Problem 2. Express answers to the following using the correct number of sig figs. Check your answers.You may be surprised!

a. 3.14 m X 3m =

b. 4.688 m / 2.0 m =

c. 3.4 m + 2.11 m + 0.8001 m =

d. (4.811 m)(3.1 m)(5 m) =

e. (3.4 m)(9.22 m) / 3.2 m =

f. (6.68 m2 / 2.2 m) - 3.4 m + 7.88 m =

g. (3.42 m2 / 2.1 m)(2.442 m / 2.10 m)(8.866 m / 2.14 m) + 4.532 m =

h. 34.5772 cm + 0.43 cm =

i. (345.2 m)(2.01m) =

j. 457.8865 mL / 3 mL =

k. 889 g - 2.886 g =

l. (45 g + 124 g) / 20. g =

m. (23,800 m)(2 m) =

3. Indicate how many sig figs are present in each of the following.

a. 671 m _____ j. 40003 m _____

b. 360 m _____ k. 0.00100 m _____

c. 059 m _____ l. 408.0 m _____

d. 609 m _____ m. 20000 m _____

e. 3040 m _____ n. 200130 m _____

f. 6009 m _____ o. 20.000 m _____

g. 3564.20 m _____ p. 0.050440 m _____

h. 0.042 m _____ q. 5744 m _____

i. 55600 m _____ r. 99100 m _____

4. When a number is converted into exponential form, only significant figures (digits) are used. Keepingthis in mind, express the following in scientific notation.

a. 381 cm ______________________ e. 6040 L ______________________

b. 80462 kg ______________________ f. 0.003400 g ______________________

c. 0.0055 g ______________________ g. 300 mL ______________________

d. 101000 mm ______________________ h. 0.52 kg ______________________

It is important to emphasize that significant figures only exist in measurements. Therefore, therules regarding significant figures only apply when you are working with measurements. At this point youneed to learn that there are three particular cases in which the rules regarding significant figures do notapply.

First, the rules do not apply when you are working with pure numbers. So, for example, if we were tomultiply 5 times 45, the answer would be: 5 X 45 = 225. These are pure numbers, they do not have units.Since they are not measurements, we do not round the answer using the rules for significant figures.


However, if we multiply 5 cm X 45 cm then the rules for sig figs do apply and the answer would be:

5 cm X 45 cm = 225 cm2 = 200 cm2 (rounded)

Second, the rules do not apply when you are dealing with definitions that relate units of measure withinthe same system - such as the metric system. For example, look at the problem below and calculate theanswer.

434.5mL X

1 L

1000 mL = ??? L

This problem involves multiplication and division. Therefore, we count the number of sig figs in each ofthe measurements. The only measurement in the problem is 434.5 mL. The ratio of 1L/1000mL is adefinition. Even though this ratio includes numbers with units, 1 liter is equal to 1000 mL by definition.Therefore, we look only at 434.5 mL. Seeing that it contains 4 sig figs, we round the answer to four sigfigs. If a ratio is used which changes units in one system of measurement to an equivalent measure inanother system of measurement, then the rules do apply. For example, calculate the answer to theproblem below.

6764 g X

1.00 pound

454 grams = ??? pounds

The ratio in the problem above converts pounds (English system) into grams (metric system).Measurements actually had to be made to determine this relationship, so the rules apply here. You cannotchange pounds into grams by definition since they belong to different systems. Following the rule formultiplication and division, the answer is rounded to 3 sig figs.

Third, the rules for significant figures do not apply to "counts," because "counts" are not measurements.For example, in the problem below, calculate the number of dozens of eggs we can obtain from 155 eggs:

8064 eggs X

1 dozen

12 eggs = ??? dozens

8064 eggs is a "count." There is no uncertainty. It means exactly 8064 eggs. Similarly, there are exactly12 eggs in a dozen. No measuring instrument is needed here, thus these cannot be measurements. Therules for sig figs do not apply. The answer is 672 dozens.

SECTION A-3. Answers to Problems

1. a. 2.4421 cm h. 42.0040 m o. 7080.0940 mg

b. 200.41 m i. 3000 mm p. 0.8 mg

c. 3.00 L j. 5.300 g q. 674 L

d. 0.10004 cm3 k. 00.0050050 mL r. 4000200.080 m

e. 0.0020 m2 l. 240 kg s. 767003 cm

f. 00.0030030 m m. 23,000.010 cm t. 97600 g

g. 108,090 cm n. 0.0060 mL u. 8740. mg

2. a. 9 m2; b. 2.3 m; c. 6.3 m; d. 70 m3; e. 9.8 m; f. 7.5 m; g. 12.3 m; h. 35.01 cm; i. 694 m2; j. 200;k. 886 g; l. 8.5 g; m. 50,000 m2

3. a. 3; b. 2; c. 2; d. 3; e. 3; f. 4; g. 6; h. 2; i. 3; j. 5; k. 3; l. 4; m. 1; n. 5; o. 5; p. 5; q. 4; r. 3

4. a. 3.81 X 102 cm; b. 8.0462 X 104 kg; c. 5.5 X 10-3 g; d. 1.01 X 105 mm; e. 6.04 X 103 Lf. 3.400 X 10-3 g; g. 3 X 102 mL; h. 5.2 X 10-1 kg


APPENDIX B

Actual Electron Arrangement Of The Elements

Sublevels ---> 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d

1 hydrogen 12 helium 23 lithium 2 14 beryllium 2 25 boron 2 2 16 carbon 2 2 27 nitrogen 2 2 38 oxygen 2 2 49 fluorine 2 2 510 neon 2 2 611 sodium 2 2 6 112 magnesium 2 2 6 213 aluminum 2 2 6 2 114 silicon 2 2 6 2 215 phosphorus 2 2 6 2 316 sulfur 2 2 6 2 417 chlorine 2 2 6 2 518 argon 2 2 6 2 619 potassium 2 2 6 2 6 120 calcium 2 2 6 2 6 221 scandium 2 2 6 2 6 2 122 titanium 2 2 6 2 6 2 223 vanadium 2 2 6 2 6 2 324 chromium 2 2 6 2 6 1 525 manganese 2 2 6 2 6 2 526 iron 2 2 6 2 6 2 627 cobalt 2 2 6 2 6 2 728 nickel 2 2 6 2 6 2 829 copper 2 2 6 2 6 1 1030 zinc 2 2 6 2 6 2 1031 gallium 2 2 6 2 6 2 10 132 germanium 2 2 6 2 6 2 10 233 arsenic 2 2 6 2 6 2 10 334 selenium 2 2 6 2 6 2 10 435 bromine 2 2 6 2 6 2 10 536 krypton 2 2 6 2 6 2 10 637 rubidium 2 2 6 2 6 2 10 6 138 strontium 2 2 6 2 6 2 10 6 239 yttrium 2 2 6 2 6 2 10 6 2 140 zirconium 2 2 6 2 6 2 10 6 2 241 niobium 2 2 6 2 6 2 10 6 1 442 molybdenum 2 2 6 2 6 2 10 6 1 543 technetium 2 2 6 2 6 2 10 6 2 544 ruthenium 2 2 6 2 6 2 10 6 1 745 rhodium 2 2 6 2 6 2 10 6 1 846 palladium 2 2 6 2 6 2 10 6 1047 silver 2 2 6 2 6 2 10 6 1 1048 cadmium 2 2 6 2 6 2 10 6 2 10 49 indium 2 2 6 2 6 2 10 6 2 10 150 tin 2 2 6 2 6 2 10 6 2 10 251 antimony 2 2 6 2 6 2 10 6 2 10 352 tellurium 2 2 6 2 6 2 10 6 2 10 4


B-1 ©1997, A.J. Girondi


53 iodine 2 2 6 2 6 2 10 6 2 10 554 xenon 2 2 6 2 6 2 10 6 2 10 655 cesium 2 2 6 2 6 2 10 6 2 10 6 156 barium 2 2 6 2 6 2 10 6 2 10 6 257 lanthanum 2 2 6 2 6 2 10 6 2 10 6 2 158 cerium 2 2 6 2 6 2 10 6 2 10 6 2 1 159 praseodymium 2 2 6 2 6 2 10 6 2 10 6 2 360 neodymium 2 2 6 2 6 2 10 6 2 10 6 2 461 promethium 2 2 6 2 6 2 10 6 2 10 6 2 562 samarium 2 2 6 2 6 2 10 6 2 10 6 2 663 europium 2 2 6 2 6 2 10 6 2 10 6 2 764 gadolinium 2 2 6 2 6 2 10 6 2 10 6 2 7 165 terbium 2 2 6 2 6 2 10 6 2 10 6 2 966 dysprosium 2 2 6 2 6 2 10 6 2 10 6 2 1067 holmium 2 2 6 2 6 2 10 6 2 10 6 2 1168 erbium 2 2 6 2 6 2 10 6 2 10 6 2 1269 thulium 2 2 6 2 6 2 10 6 2 10 6 2 1370 ytterbium 2 2 6 2 6 2 10 6 2 10 6 2 1471 lutetium 2 2 6 2 6 2 10 6 2 10 6 2 14 172 hafnium 2 2 6 2 6 2 10 6 2 10 6 2 14 273 tantalum 2 2 6 2 6 2 10 6 2 10 6 2 14 374 tungsten 2 2 6 2 6 2 10 6 2 10 6 2 14 475 rhenium 2 2 6 2 6 2 10 6 2 10 6 2 14 576 osmium 2 2 6 2 6 2 10 6 2 10 6 2 14 677 iridium 2 2 6 2 6 2 10 6 2 10 6 2 14 778 platinum 2 2 6 2 6 2 10 6 2 10 6 1 14 979 gold 2 2 6 2 6 2 10 6 2 10 6 1 14 1080 mercury 2 2 6 2 6 2 10 6 2 10 6 2 14 1081 thallium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 182 lead 2 2 6 2 6 2 10 6 2 10 6 2 14 10 283 bismuth 2 2 6 2 6 2 10 6 2 10 6 2 14 10 384 polonium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 485 astatine 2 2 6 2 6 2 10 6 2 10 6 2 14 10 586 radon 2 2 6 2 6 2 10 6 2 10 6 2 14 10 687 francium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 188 radium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 289 actinium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 190 thorium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 291 protactinium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 2 192 uranium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 3 193 neptunium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 4 194 plutonium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 695 americium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 796 curium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 7 197 berkelium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 8 198 californium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 1099 einsteinium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 11100 fermium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 12101 mendelevium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 13102 nobelium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14103 lawrencium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 1104 unnilquadium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 2105 unnilpentium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 3106 unnilhexium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 4107 unnilseptium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 5108 unniloctium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 6109 unnilennium 2 2 6 2 6 2 10 6 2 10 6 2 14 10 6 2 14 7Sublevels ---> 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f 6d


Appendix B. cont.-"Core Notation" of 100 Elements According to the Diagonal Rule

1 H 1s1 51 Sb [Kr]5s24d105p3

2 He 1s2 52 Te [Kr]5s24d105p4

3 Li [He]2s1 53 I [Kr]5s24d105p5

4 Be [He]2s2 54 Xe [Kr]5s24d105p6

5 B [He]2s22p1 55 Cs [Xe]6s1

6 C [He]2s22p2 56 Ba [Xe]6s2

7 N [He]2s22p3 57 La [Xe]6s24f18 O [He]2s22p4 58 Ce [Xe]6s24f29 F [He]2s22p5 59 Pr [Xe]6s24f310 Ne [He]2s22p6 60 Nd [Xe]6s24f411 Na [Ne]3s1 61 Pm [Xe]6s24f512 Mg [Ne]3s2 62 Sm [Xe]6s24f613 Al [Ne]3s23p1 63 Eu [Xe]6s24f714 Si [Ne]3s23p2 64 Gd [Xe]6s24f815 P [Ne]3s23p3 65 Tb [Xe]6s24f916 S [Ne]3s23p4 66 Dy [Xe]6s24f10

17 Cl [Ne]3s23p5 67 Ho [Xe]6s24f11

18 Ar [Ne]3s23p6 68 Er [Xe]6s24f12

19 K [Ar]4s1 69 Tm [Xe]6s24f13

20 Ca [Ar]4s2 70 Yb [Xe]6s24f14

21 Sc [Ar]4s23d1 71 Lu [Xe]6s24f145d1

22 Ti [Ar]4s23d2 72 Hf [Xe]6s24f145d2

23 V [Ar]4s23d3 73 Ta [Xe]6s24f145d3

24 Cr [Ar]4s23d4 74 W [Xe]6s24f145d4

25 Mn [Ar]4s23d5 75 Re [Xe]6s24f145d5

26 Fe [Ar]4s23d6 76 Os [Xe]6s24f145d6

27 Co [Ar]4s23d7 77 Ir [Xe]6s24f145d7

28 Ni [Ar]4s23d8 78 Pt [Xe]6s24f145d8

29 Cu [Ar]4s23d9 79 Au [Xe]6s24f145d9

30 Zn [Ar]4s23d10 80 Hg [Xe]6s24f145d10

31 Ga [Ar]4s23d104p1 81 Tl [Xe]6s24f145d106p1 32 Ge [Ar]4s23d104p2 82 Pb [Xe]6s24f145d106p2

33 As [Ar]4s23d104p3 83 Bi [Xe]6s24f145d106p3

34 Se [Ar]4s23d104p4 84 Po [Xe]6s24f145d106p4

35 Br [Ar]4s23d104p5 85 At [Xe]6s24f145d106p5

36 Kr [Ar]4s23d104p6 86 Rn [Xe]6s24f145d106p6

37 Rb [Kr]5s1 87 Fr [Rn]7s1

38 Sr [Kr]5s2 88 Ra [Rn]7s2

39 Y [Kr]5s24d1 89 Ac [Rn]7s25f140 Zr [Kr]5s24d2 91 Pa [Rn]7s25f342 Mo [Kr]5s24d4 92 U [Rn]7s25f443 Tc [Kr]5s24d5 93 Np [Rn]7s25f544 Ru [Kr]5s24d6 94 Pu [Rn]7s25f645 Rh [Kr]5s24d7 95 Am [Rn]7s25f746 Pd [Kr]5s24d8 96 Cm [Rn]7s25f847 Ag [Kr]5s24d9 97 Bk [Rn]7s25f948 Cd [Kr]5s24d10 98 Cf [Rn]7s25f10

49 In [Kr]5s24d105p1 99 Es [Rn]7s25f11

50 Sn [Kr]5s24d105p2 100 Fm [Rn]7s25f12


H Be B C N O F Ne

Group: 1A 2A 3A 4A 5A 6A 7A 8A

Dot Notations of Elements in Groups 1A - 8A


APPENDIX C

SECTION C.1 Freezing Point Depression And Boiling Point Elevation

As you should already know, the presence of a solute in a solution raises the boiling point. Themost scientific explanation for the increase in boiling point is that the solute tends to lower the vaporpressure of the solvent. That seems kind of vague, right? Maybe the graphs shown below will help. Thesolid lines represent the vapor pressure of pure water, while the dotted lines represent the vapor pressureof water in a solution. Note that the vapor pressure at any given temperature is lower when water iscombined with a solute in a solution. Also note that in a concentrated solution, the vapor pressure isdepressed more than it is in a dilute solution.

You will recall that water will boil only when its vapor pressure equals the atmospheric pressure (760mm Hg at sea level). Note that the vapor pressure of pure water equals atmospheric pressure at 100oC;thus, 100oC is the boiling point of water at sea level. (See T1 on the graphs.) However, note that thetemperature of the water in a solution has to go higher before its vapor pressure will equal atmosphericpressure. (See T2 on the graphs below.) Therefore, the boiling point of the water in the solution is above100oC. Also note that the boiling point of the water in the concentrated solution will be higher than that ofthe water in the dilute solution.

760 mm

T1 T2

Concentrated Solution

V.P.

Temperature

760 mm

T1 T2

Dilute Solution

V.P.

Temperature

A

B

A = vapor pressure of pure water at T1

B = vapor pressure of water in a solution at T1

vapor pressureof pure water

vapor pressure of water in a solution

A

B

A = vapor pressure of pure water at T1

B = vapor pressure of water in a solution at T1

C

C

A = boiling point of pure waterC = boiling point of water in a solution

A = boiling point of pure waterC = boiling point of water in a solution

C-1 ©1997, A.J. Girondi

Above 0oC water is a liquid rather than a solid because the molecules have enough kinetic energy(energy of motion) to overcome the attractive forces between them that could "lock" them into a nearlyfixed position. At 0oC the molecules have lost enough energy so that the attractive forces between themmove the molecules into a solid crystalline arrangement. The presence of a solute in the water interfereswith this process, and the result is a lower freezing point. You should have already completed an activity inwhich you determined the freezing point of a salt water solution. You should have noted that an ice bathof salt water is colder than an ice bath of pure water. Now, why does the addition of salt make ice watercolder? It seems kind of mysterious, doesn't it?

In ice water (at 0oC) the ice is melting and the water is freezing. This is because 0oC is both thefreezing point of water and the melting point of ice. As the ice melts, it absorbs heat from the water (anendothermic process). As the water freezes, the heat in it is released to the water (an exothermicprocess). It all makes sense if you keep in mind that an ice water bath represents an equilibrium condition.That's the secret to understanding this phenomenon. When we say that the system is at equilibrium, wemean that the heat is being released to and removed from the water at equal rates; thus, the temperatureremains constant at 0oC. However, when you add salt the equilibrium is upset. The salt will not affect therate at which the ice melts, but it will slow down the rate at which the water freezes. The particles from thesalt (solute) get in between the water (solvent) molecules and weaken the attraction between the watermolecules. This causes the freezing rate of the water to slow down. Thus, the melting rate of the ice isfaster than the freezing rate of the water. The salt has allowed the endothermic process (melting) to gofaster than the exothermic process (freezing). That means that heat is being removed from the waterfaster than it is being replaced. The result is that the temperature of the water goes down. When all of theice is consumed, the melting can no longer continue, and the temperature drop will cease.

water freezing

exothermicice

ice water melting

endothermic

water freezing (slower)

exothermicice

ice water melting (faster)

endothermic

Ice Water (equal rates) Salt Water (unequal rates)

Homemade ice cream requires that the cream be frozen in a mixture of rock salt and ice. Ice wateralone will not freeze the cream. If you know the concentration of a solution, you can actually calculate howmuch the boiling point will go up of how much the freezing point will go down. However, you cannot useconcentrations expressed as molarity (M). Remember that molarity (M) tells you nothing about the amountof solvent in a solution. The units for molarity are moles of solute per liter of solution. Solvent is not evenmentioned in the definition. Molarity is commonly used in chemistry because we usually don't care aboutthe amount of solvent in a solution. We are always concerned about the amount of solute. However,freezing point depression and boiling point elevation are properties that depend on the ratio of soluteparticles to solvent particles in a solution. Such properties are known as colligative properties. Osmoticpressure, which you may have read about in biology class, is another example of a colligative property ofsolutions. To calculate the change in the freezing point or boiling point of a solution, you will need to usea unit of concentration that contains information about both the solute and the solvent. Molality (m)contains this information. Molality (m) is defined as moles of solute per kilogram of solvent. A lowercase"m" represents molality.

1 mole solute

1 kg solvent1 m =

First, let's try some problems that involve only the concept of molality, and then we will look at problemswhich reveal how molality is used to calculate changes in freezing and boiling points.


Sample Problem: What is the molal concentration of a solution which contains 45 grams of NaOHdissolved in 400. mL of water?

First, note that the amount of water is given in the problem, whereas the amount of solution would havebeen given if we were working with molarity. We want to end up with units of moles NaOH/Kg H2O. Sincewe want to end with a ratio of information about solute divided by information about solvent, let's start withthe information about solute divided by the information about solvent: 45 g NaOH/400 mL H2O.

45 g NaOH

400. mL H 2O X

?

? X

?

? X

?

? =

?? moles NaOH

kg H2O

Remember that 1 g H2O = 1 mL H2O, and that 1 Kg H2O = 1000 g H2O. Complete the problem in thespace below. The correct answer is 2.8 moles NaOH /1 Kg H2O,or 2.8 m. Solve the problem after youcomplete the set-up and see if you can obtain this answer.

45 g NaOH

400. mL H 2O X

X

X

=

moles NaOH

kg H2O

Now try the problems below.

Problem 1. What is the molal concentration of a solution which contains 188 grams of BaCl2 and 250.mL of water?

__________ m

Problem 2. What is the molality of a solution if 45.0 grams of HCl are dissolved in 500. mL of water?

__________ m

Problem 3. If 0.360 kilogram of C12H22O11 is dissolved in 788 grams of water, what is the molalconcentration?

__________ m


Problem 4. How many grams of KOH must be added to 0.45 kg of water to make a 0.33 m (molal)solution of KOH?

__________ g KOH

Problem 5. How many grams of water are needed to make a 1.200 m solution of NaCl which contains75.00 grams of NaCl?

__________ g H2O

There are formulas which can be used to determine the amount that a solute lowers a freezing pointor raises a boiling point:

∆Tƒ = (kƒ ) (m) (i) and ∆Tb = (kb) (m) (i)

∆Tƒ and ∆Tb stand for the change in the freezing point and the change in the boiling point, respectively.The symbols kƒ and kb represent the molal freezing-point constant and the molal boiling-point constant,respectively. These quantities are experimentally determined, and are found in reference books. Like allscientific constants, they are present in the equations to "make them work." The lowercase "m"represents molality, while the "i" is called the activity coefficient. In simple terms, "i" represents thenumber of moles of particles which are formed when one mole of a particular solute dissolves.

Electrolytes are substances which conduct an electric current when dissolved in solvents, whilenonelectrolytes are substances which do not conduct an electric current when dissolved in solvents.Many ionic and polar covalent compounds are electrolytes, while nonpolar covalent substances arenonelectrolytes. For all nonelectrolytes, "i" always has a value of 1. For electrolytes, the value of "i"depends on the number of particles the substance breaks into (dissociates into) when dissolved insolution. (Remember that polyatomic ions do not break apart in solution.) For example, let's consider thefour examples below.

When one mole of table sugar molecules (a nonelectrolyte with formula C12H22O11) dissolves inwater, it forms only one mole of particles. This is true of all nonelectrolytes. One mole of anynonelectrolyte forms one mole of particles, since nonelectrolytes do not split up into ions when theydissolve.

nonelectrolyte: C12H22O11(s) -------> C12H22O11(aq) i = 1

When one mole of the electrolyte NaOH dissolves in water, it forms two moles of particles (1 mole of Na1+

ions and one mole of OH1- ions).

electrolyte: NaOH(s) -------> Na1+(aq) + OH1-(aq) i = 2


One mole of the electrolyte Sr(NO3)2 will form three moles of particles when it dissolves in water:

electrolyte: Sr(NO3)2(s) -------> Sr2+(aq) + 2 NO31-(aq) i = 3

The electrolyte AlCl3 will form four moles of particles per one mole of solute. Examine the equation below.

electrolyte: AlCl3(s) -------> Al3+(aq) + 3 Cl1-(aq) i = 4

This is important because it is the number of moles of solute particles in solution, rather than just the number of moles of solute, which are important when you are studying colligative properties. If you are working with a nonelectrolyte, "i" is always equal to one, because nonelectrolytes are composed of molecules which do not break up into smaller parts in solution. However, if you are working with an electrolyte, you must look at the formula to determine how many parts the substance breaks into and to determine the value of "i." Therefore, you must either know or be told whether or not the substance you are working with is an electrolyte. There are no units generally assigned to "i." In addition, "i" is not a measurement, so its value will not affect the number of significant digits in the answers to problems.

Problem 6. For the electrolytes listed below, show how they break into particles, and determine the value of "i" for each substance. Be sure to show the charges on ions which form.

a. Ca(OH)2(s) -----> ________________________________ i = ________

b. KMnO4(s) -----> ________________________________ i = ________

c. GaF3(s) -----> ________________________________ i = ________

d. MgBr2(s) -----> ________________________________ i = ________

e. LiCl(s) -----> ________________________________ i = ________

f. Na3PO4(s) -----> ________________________________ i = ________

g. (NH4)2SO4(s) -----> ________________________________ i = ________

As for the constants, for water Kƒ has a value of 1.86 and kb has a value of 0.52. Other solvents have different values for these constants. These values may be used only for water solutions. Constants in science can have strange units. The unit for these is (kg solute)(oC)/(moles solute). Why that, you ask? Well, we want ∆Tƒ and ∆Tb to have values expressed in oC. Look at the formulas for ∆Tƒ and ∆Tb given previously and figure it out for yourself.

For all water solutions: kƒ = 1.86 and kb = 0.52

Sample Problem: What are the freezing and boiling temperatures of a 0.10 m solution of NaOH (an electrolyte)?

We begin by calculating the change in the freezing point of water:

∆Tƒ = (kƒ ) (m) (i) so, ∆Tƒ = (1.86)(0.10)(2) = 0.37oC

Since the freezing point goes down, we subtract ∆Tƒ from the normal freezing point of water:

0oC - 0.37oC = -0.37oC = Tƒ = freezing point of the solution


Next, we calculate the change in the boiling point of water:

∆Tb = (kb)(m)(i) so, ∆Tb = (0.52)(0.10)(2) = 0.10

Since the boiling point goes up, we add ∆Tb to the normal boiling point of water:

100oC + 0.10oC = 100.10oC = Tb = boiling point of solution

Problem 7. Now, determine the freezing and boiling points of the solutions in problems 1, 2, and 3 which you solved earlier in this appendix, keeping in mind that BaCl2 and HCl in problems 1 and 2 are electrolytes, while C12H22O11 in problem 3 is a nonelectrolyte.

a.

Tƒ = __________oC; Tb = __________oCb.

Tƒ = __________oC; Tb = __________oCc.

Tƒ = __________oC; Tb = __________oC

Problem 8. The label fell off of a bottle of a 0.20 molal water solution in the laboratory, and an attempt isbeing made to identify the solute. It is known to be either KBr, CaCl2, or GaI3. Tests reveal that thesolution is an electrolyte. Its freezing point is -1.1oC. What is the identity of the unknown solid? Show anycalculations, and explain your conclusion.

Calculations:

Explanation: ____________________________________________________________________

______________________________________________________________________________

Problem 9. A car owner puts 301 grams of ethylene glycol, C2H6O2, in his car's radiator for every 750.grams of water in it. This is a commonly used antifreeze mixture. How cold will it have to get before thisguy will have to worry about a frozen cooling system? (Ethylene glycol is a nonelectrolyte.)

_________oC


SECTION C.2 Answers to Problems

1. 3.61 m

2. 2.47 m

3. 1.33 m

4. 8.3 g

5. 1069 g

6. a. Ca(OH)2(s) -----> Ca2+(aq) + 2 OH1-(aq); i = 3b. KMnO4(s) -----> K1+(aq) + MnO41-(aq); i = 2c. GaF3(s) -----> Ga3+(aq) + 3 F1-(aq); i = 4d. MgBr2(s) -----> Mg2+(aq) + 2 Br1-(aq); i = 3e. LiCl(s) -----> Li1+(aq) + Cl1-(aq); i = 2f. Na3PO4(s) -----> 3 Na1+(aq) + PO43-(aq); i = 4g. (NH4)2SO4(s) -----> 2 NH41+(aq) + SO42-(aq); i = 3

7. a. Tƒ = -20.1oC; Tb = 105.6oCb. Tƒ = -9.19oC; Tb = 102.6oCc. Tƒ = -2.47oC; Tb = 100.69oC

8. molality = 0.20m; i = 2.96 which rounds to 3 since "i" is a whole number; Since i = 3, the solute must be CaCl2 : CaCl2(s) ----> Ca2+(aq) + 2 Cl1-(aq)


SECTION C.3 Student Notes


APPENDIX D

SPECIFIC HEAT, HEAT CAPACITY, AND TEMPERATURETHE "SWIMMING POOL" ANALOGY

Let's begin with a few definitions. Temperature and heat are not the same. Temperature is ameasure of the average kinetic energy of the molecules of a system, while heat is a measure of the totalkinetic energy of a system. You can also think of temperature as a measure of the ability of a system totransfer heat to another system. Heat flows from high temperature systems to low temperature systems.

Specific heat is defined as the number of calories or joules of heat required to raise thetemperature of 1 gram of substance by 1oC. Therefore, specific heat has units of cal/g.oC or J/g.oC.Different substances have different specific heats. The larger the specific heat of a substance, the greaterthe number of calories or joules required to raise the temperature of 1 gram of the substance by 1oC. So,if a substance has a high specific heat, then a lot of heat is required to raise its temperature. Thetemperatures of substances with smaller specific heats can be increased a great deal with smaller amountsof heat.

As an analogy, let's consider two swimming pools - a large one and a small one. Think of the largepool as a substance with a large specific heat value, and of the small pool as a substance with a smallspecific heat value. Think of the water level in the pools as temperature and of the amount of water in thepools as heat. Now add an equal amount of water (heat) to each pool. The water level (temperature) of thesmall pool has gone up more than that of the large pool. Therefore, the temperatures of substances withsmall specific heat values go up more than those of substances with large specific heat values when equalamounts of heat are gained.

Large Pool(large specific heat)

Small Pool(small specific heat)

temp

temp

HEATHEAT

If a car is exposed to the sun on a frosty morning and begins to get warmer, the frost on the car willmelt off of the metal car body before it melts off of the glass windshield. Assume that both the metal andthe glass absorb equal amounts of heat, but that the temperature of the metal goes up more becausemetal has a smaller specific heat than glass does. Since temperature is a measure of the ability to transferheat, the metal is more able to transfer heat to the frost than the glass is. That's why the frost lingers onthe windshield and frequently has to be scraped off.

Molar heat capacity is very similar to specific heat. The only difference is that molar heat capacityrefers to 1 mole of a substance, while specific heat refers to 1 gram of a substance. The unit for molar heatcapacity is cal/mole oC or J/mole oC. Since the glass on the car has a higher specific heat than the metal, italso has a higher heat capacity than the metal does. Molar heat capacity can also be interpreted as theamount of heat that 1 mole of a substance can hold for every degree of temperature. For example, 1 moleof a substance with a molar heat capacity of 80 J/mole oC can hold 80 joules of heat for each 1 oC. If itstemperature is 10oC, then 1 mole of the substance can hold 800 joules of heat. If 1 moles of anothersubstance has a molar heat capacity of 30 J/mole oC, then at 10oC it can hold only 300 joules of heat. Inother words, the higher the heat capacity of a substance, the more heat it can hold at a given temperature.So, as the car sits out in the sun, the metal body will get hotter faster. But, if the temperatures of the metaland glass eventually equalize, the glass will have more heat in it than the metal will. Glass has a greatercapacity to hold heat (heat capacity) than the metal does.

D-1 ©1997, A.J. Girondi

Summary:

1. With a small heat capacity, a substance warms quickly when heated, but can't hold much heat. Addwater to the small pool and the water level (temperature) goes up fast, but it can't hold much water (heat).

2. With a large heat capacity, a substance warms slowly when heated, but can hold a lot of heat. Add waterto the large pool and the water level (temperature) goes up slowly, but it can hold lots of water (heat).

Note: Good cookware is made of materials with high heat capacities or specific heats. They take a littlelonger to get hot, but the large amount of heat that they hold tends to keep cooking temperatures moreeven. (The water level in the large pool tends to fluctuate less than that of the small pool.)

- End of Appendix D

D-2 ©1997, A.J. Girondi

APPENDIX E

SECTION E.1 The Solubility Product Constant

You have probably already studied equilibrium systems that involve chemical changes. However,

equilibrium systems can also involve physical changes. For example, consider the system illustrated bythe equation below:

NaCl(s) <===> Na1+(aq) + Cl1-(aq)

This system involves the physical processes of dissolving and crystallization. A solution in which thesetwo processes are in equilibrium is said to be {1}_________________. Notice, too, that there is a solidinvolved in this equilibrium. You will recall from your previous studies that solids and pure liquids are neverincluded in equilibrium expressions. With that in mind, examine the equilibrium expression for thissystem:

equilibrium expression = [Na1+] [Cl1-]

Notice that the solid NaCl has not been included in the expression. When the equilibrium system involvesthese physical processes of dissolving and crystallization, the constant is given a different name. Insteadof calling it the equilibrium constant, Keq, it is called the solubility product constant and is given the symbol,Ksp. Because solids are not included, Ksp, expressions have no denominator! So,

Ksp = [Na1+] [Cl1-]

When an ionic solid is dissolved in water in sufficient quantity to saturate the solution, anequilibrium is established between the ions in the saturated solution and the molecules of excess solid inthe container: NaCl(s) <===> Na1+(aq) + Cl1-(aq) Adding more solid to a solution which is alreadysaturated does not alter the amount of solid that is dissolved. Thus, adding more NaCl(s) to the systemabove will not change the concentrations of Na1+ or Cl1- ions in the solution. Since the amount of solidNaCl does not affect the equilibrium, it is omitted from the equilibrium expression. This is often the casewhen solids are involved in equilibrium systems.

Problem 1. Write the Ksp expressions for the systems defined by the four equations below. Rememberthat coefficients in the equations become exponents in the equilibrium expressions.

a. AgCl(s) <===> Ag1+(aq) + Cl1-(aq) Ksp =

b. CdS(s) <===> Cd2+(aq) + S2-(aq) Ksp =

c. PbCl2(s) <===> Pb2+(aq) + 2 Cl1-(aq) Ksp =

d. Ag3PO4(s) <===> 3 Ag1+(aq) + PO43-(aq) Ksp =

The size of the Ksp value indicates just how soluble a particular salt is in water. The larger the Ksp value, themore soluble the salt. However, it is wise to compare only salts with "analogous" Ksp expressions. Forexample, of the four salts listed in problem 1, the Ksp expressions of AgCl and CdS are analogous. In eachcase, two different ion concentrations are involved and they have comparable exponents. Theexpressions for PbCl2 and Ag3PO4 are not analogous to any of the other expressions listed.

The Ksp values of some selected salts are listed in Table E.1. The Ksp expressions in Table E.1are all analogous.

E-1 ©1997, A.J. Girondi

Table E–1Ksp Values of Selected Salts

Salt Equilibrium Ksp Expression Ksp Value

AgCl(s) <===> Ag1+(aq) + Cl1-(aq) Ksp = [Ag1+] [Cl1-] 1.7 X 10-10

CaCO3(s) <===> Ca2+(aq) + CO32-(aq) Ksp = [Ca2+] [CO32-] 8.8 X 10-9

PbCrO4(s) <===> Pb2+(aq) + CrO42-(aq) Ksp = [Pb2+] [CrO42-] 2.0 X 10-14

ZnS(s) <===> Zn2+(aq) + S2-(aq) Ksp = [Zn2+] [S2-] 1.0 X 10-23

BaSO4(s) <===> Ba2+(aq) + SO42-(aq) Ksp = [Ba2+] [SO42-] 1.4 X 10-9

Which salt listed in Table E.1 is most soluble in water? {2}______________________

Which salt listed in Table E.1 is least soluble in water? {3}______________________

The value of Ksp can be calculated from solubility data. In Chapter 16 you calculated thesolubilities of various substances. Now we will use those solubilities to calculate Ksp values. Study thesample problem below. Use the procedure shown to solve the problems that follow.

Sample Problem: The solubility of AgCl in water at 25oC is 1.26 X 10-5 mole / liter. Calculate Ksp.

Solution:

a. Write the equilibrium expression: AgCl(s) <===> Ag1+(aq) + Cl1-(aq)

b. Write the Ksp expression: Ksp = [Ag1+] [Cl1-]

c. Examine the mole ratio in the equation. In this case it is 1:1:1. So, if 1.26 X 10-5 mole of AgCl dissolvesper liter of solution, then it would form 1.26 X 10-5 moles per liter of Ag1+ ions and the same number of Cl1-

ions. Therefore, substituting into the Ksp expression:

Ksp = [1.26 X 10-5 ] [1.26 X 10-5 ] = 1.59 X 10-10

Problem 2. Calculate Ksp for PbS given that the solubility of PbS is water at 25oC is 7.07 X 10-5 M.

Problem 3. Calculate Ksp for PbCl2 given that the solubility of PbCl2 at 25oC is 1.59 X 10-3 M. (Becareful. The ratio here is not 1:1:1.)


The Ksp value of a substance can also be used to calculate the concentration of one ion if youknow the concentration of the other ion in a saturated solution. Study the next two sample problems.

Sample Problem: The Ksp value of CaCO3 is 5.00 X 10-9 at 25oC. Calculate [Ca2+] in a saturatedsolution in which [CO32-] = 0.400 M.

Solution:

a. Write the equilibrium expression: CaCO3(s) <===> Ca2+(aq) + CO32-(aq)

b. Write the Ksp expression: Ksp = [Ca2+] [CO32-]

c. Substitute the known information into the Ksp expression: 5.00 X 10-9 = [Ca2+] [CO32-]

[Ca2+] =

5.00 X 10-9 M2

0.400 M = 1.25 X 10 -8 Md. Solve for the unknown quantity:

Sample Problem: The K sp value of BaF2 is 2.4 X 10-5 at 25oC. Find [F1-] in a saturated solution ofBaF2 if [Ba2+] = 4.8 X 10-2 M.

Solution:

a. Write the equilibrium expression: BaF2(s) <===> Ba2+(aq) + 2 F1-(aq)

b. Write the Ksp expression: Ksp = [Ba2+] [F1-]2

c. Substitute the known information into the Ksp expression: 2.4 X 10-5 = [4.8 X 10-2] [F1-]2

d. Solve for the unknown quantity:

[F1− ]2 = 2.4 X 10-5 M3

4.8 X 10 -2 M = 5.0 X 10-4 M2

[F1− ] = 2.2 X 10 -2 M

Now try the problems below.

Problem 4. The Ksp of barium carbonate, BaCO3, at 16oC is 7.0 X 10-9. Calculate the concentration ofbarium ions, given that the equilibrium mixture contains 0.040 M of carbonate ions, CO32-.

Problem 5. The Ksp of silver sulfate, Ag2SO4 is 1.2 X 10-5 at 25oC. Find the [Ag1+] in a solution inwhich the [SO42-] = 6.2 X 10-4.


Ksp values are useful in much the same way as Keq values are. NaCl has a much larger Ksp valuethan AgCl. From this information we can conclude that NaCl is much {4}__________ soluble than AgCl.

ACTIVITY E.2 Qualitative Analysis of Solutions

When solutions containing two different soluble salts are mixed together, a chemical reaction mayoccur. As you saw in a earlier chapter, a chemical reaction may be evidenced by a change in color,evolution of a gas, a change in temperature, or by the formation of a precipitate. In cases where aprecipitate forms, the amount formed is directly related to the size of the Ksp value of the precipitatingsolid. In this activity, you will be seeing a number of chemical reactions as you mix various solutionstogether. Substances which have very small Ksp values are not very soluble. So, when you mix ionstogether which result in the formation of such a substance, most of the substance will take the form of anundissolved solid precipitate.

In this activity, you will be given six labeled solutions. You will mix all of the solutions (in pairs) witheach other, making careful observations of any chemical reactions that occur. You will then be given four"unknown" solutions. Three of the unknowns will be identical to three of the original solutions. The fourthunknown will contain a combination of two of the original solutions. Your task will be to establish theidentity of each unknown. When the purpose of an experiment is to identify the substance(s) present inunknowns, the procedure is called qualitative analysis. By mixing a little of each of the unknowns with thesix known solutions, you should be able to determine their identities. You will also be testing each of theknown and unknown solutions with litmus paper to identify each solution as an acid or base. (You will bestudying acids and bases shortly, if you haven't already.)

(Note: some of the solutions are naturally blue, so blue food coloring was added to the others solutionsso that they all look the same.)

Obtain a dropping plate and the six known solutions. Before you do any mixing, test eachsolution with litmus paper by adding a drop of each solution to a piece of red litmus paper and anotherdrop to a piece of blue litmus paper. If the red paper turns blue, the solution is a base. If the blue paperturns red, the solution is an acid. If no change occurs with either paper, then the solution is neutral. Enterthe results in Table E.1. Mix the solutions in all possible combinations by adding about 5 drops of onesolution to a well of the dropping plate, and then adding 5 drops of another solution to the same well.Observe the result and place your data in the proper place in Table E.1. Repeat this procedure until youhave completed all possible pair combinations of the six solutions.

Next, clean and dry your dropping plate. Obtain four small, clean test tubes and label them withnumbers 1 through 4. Place them in a beaker or test tube rack and give them to your instructor. He will filleach tube with an unknown solution for you to identify. Tubes 1 through 3 will each contain one of the sixsolutions which you studied previously. Tube four will contain a mixture of two of those six solutions. Bymixing and testing these four unknown solutions as you did with the known solutions previously, youmust identify the solution in each of the four tubes. Record your observations and results in Table E.2.

SECTION E.3 Answers To Questions and Problems

Questions: {1} saturated; {2} CaCO3; {3} ZnS; {4} more

Problems:

1. a. [Ag] [Cl1-]; b. [Cd2+] [S2-]; c. [Pb2+] [Cl1-]2; d. [Ag1+]3[PO43-]2. 5.00 X 10-9; 3. 1.6 X 10-8; 4. 1.75 X 10-7; 5. 1.4 X 10-1


Table E.1Qualitative Analysis of Six Known Solutions

NaHCO3 HCl BaCl2 CuSO4 KNO3 Cu(NO3)2 Acid or Base

NaHCO3

HCl

BaCl2

CuSO4

KNO3

Cu(NO3)2


Table E.2Qualitative Analysis of Four Unknown Solutions

NaHCO3 HCl BaCl2 CuSO4 KNO3 Cu(NO3)2 Acid or Base

Unknown #1

Unknown #2

Unknown #3

Unknown #4

Identities of Unknowns: #1: ________________

#2: ________________

#3: ________________

#4: ________________ and ________________

End of Appendix E


APPENDIX F

SECTION F.1 Use of Inverse Logarithms in Acid–Base Problems

Calculation of pH, given [H1+], is rather straightforward. However, doing the reverse (calculating[H1+], given pH) is a bit trickier, because it involves the use of inverse logarithms which are also known asantilogarithms. An in-depth discussion of "antilogs" is beyond the purpose of this discussion. Let's justrecognize that since an antilog is the inverse of a log, when you multiply a "log" by an "antilog," the resultin one: (antilog) X (log) = 1.

Antilogs are numbers, whereas logs are exponents. For example, consider the expressionbelow.

102 = 100

log

antilogbase

The 10 is called the base, the exponent 2 is the logarithm, and the number 100 is the antilogarithm or"antilog." An antilog is the number you get when you raise ten to a certain power.

Inverse logs (antilogs) can be found quickly using a scientific calculator. If your calculator has an[inverse] key, you may be able to get an inverse log by pressing the [inverse] key followed by the [log]key. You can also find antilogs by using the [10x] key or the [yx] key.

Let's try to find an antilog on your calculator. We will try to find the antilog of 2. Another way ofsaying this is, "what number do you get when 10 is raised to the 2nd power?" We already know that theanswer is 100, but let's try to find it using the calculator. If you have an [inverse] key, start by pressing thenumber 2. Then press the [inverse] key followed by the [log] key. If you got an answer of 100, you havesucceeded. If you got something else, check the instruction manual for your calculator to determine thecorrect procedure for finding antilogs.

To find the antilog of 2 using the [10x] key, first enter the number 2, then press the [10x] key. Thenumber 100 should appear in the display. (To use the [10x] key on some calculators, you may have topress the [second function] key first, followed by the [10x] key.

To find the antilog of 2 using the [yx] key, first enter the number 10, which is the base. Next, pressthe [yx] key followed by the number 2. Finally, press the [=] key. The display should read 100. You canraise any base to any power using the [yx] key.

Problem 1. Find the antilogs of the following logs.

a. 4 ___________ e. 67 ___________

b. 3.540 ___________ f. 0.24 ___________

c. 0.83 ___________ g. 1.09 ___________

d. –2.13 ___________ h. –0.88 ___________

F-1 ©1997, A.J. Girondi

According to part "g" of problem 1, if you want to get 12.3, you must raise 10 to the 1.09 power.Supposedly: 101.09 = 12.3 Let's see if that's correct. Enter the number 12.3 into yourcalculator. Now if you press the [log] key, you will find the power to which 10 must be raised to give you12.3. Try it. You should get 1.09 (rounded).

Now Let's try a problem involving pH. Suppose you know that the pH of a solution is 5.9, and youwant to find [H1+] for the solution. We know that: pH = - log [H1+], so let's substitute the 5.9 into theequation: 5.9 = - log [H1+]. To solve for [H1+], we need to isolate it on one side of the equation. First,multiply both sides by -1, which yields: -5.9 = log [H1+]. Next, multiply both sides by antilog:

antilog -5.9 = antilog (log [H1+])

All that is left to do is to find the antilog of -5.9. The answer you should get is 1.3 X 10-6 (rounded to 2 significant figures). Now try the problems below.

Problem 2. What is the [H1+] of a solution that has a pH of 8.22?

Problem 3. What is the [H1+] of a solution that has a pH of 1.23?

Problem 4. What is the [OH1-] of a solution which has a pH of 4.2? (In solving this problem, use the equation Kw = [H1+] [OH1-], and recall that Kw has a constant value of 1.0 X 10-14.)

Problem 5. What is the [OH1-] of a solution which has a pH of 9.11? (Again, use the Kw equation.)

SECTION F.2 Answers to Problems

1. a. 10,000; b. 3467; c. 6.8; d. 0.00741; e. 1 X 1067; f. 1.7; g. 12.3; h. 0.132. 6.03 X 10-9

3. 5.89 X 10-2

4. 1.6 X 10-10

5. 1.29 X 10-5

End of Appendix F

F-2 ©1997, A.J. Girondi

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