-
Notifications
You must be signed in to change notification settings - Fork 2
/
s08-02-solution-concentrations.html
1358 lines (1321 loc) · 97.7 KB
/
s08-02-solution-concentrations.html
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
882
883
884
885
886
887
888
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945
946
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<link href="shared/bookhub.css" rel="stylesheet" type="text/css">
<title>Solution Concentrations</title>
</head>
<body>
<div id=navbar-top class="navbar">
<div class="navbar-part left">
<a href="s08-01-aqueous-solutions.html"><img src="shared/images/batch-left.png"></a> <a href="s08-01-aqueous-solutions.html">Previous Section</a>
</div>
<div class="navbar-part middle">
<a href="index.html"><img src="shared/images/batch-up.png"></a> <a href="index.html">Table of Contents</a>
</div>
<div class="navbar-part right">
<a href="s08-03-stoichiometry-of-reactions-in-.html">Next Section</a> <a href="s08-03-stoichiometry-of-reactions-in-.html"><img src="shared/images/batch-right.png"></a>
</div>
</div>
<div id="book-content">
<div class="section" id="averill_1.0-ch04_s02" condition="start-of-chunk" version="5.0" lang="en">
<h2 class="title editable block">
<span class="title-prefix">4.2</span> Solution Concentrations</h2>
<div class="learning_objectives editable block" id="averill_1.0-ch04_s02_n01">
<h3 class="title">Learning Objective</h3>
<ol class="orderedlist" id="averill_1.0-ch04_s02_l01">
<li>To describe the concentrations of solutions quantitatively.</li>
</ol>
</div>
<p class="para editable block" id="averill_1.0-ch04_s02_p01">All of us have a qualitative idea of what is meant by <em class="emphasis">concentration</em>. Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. In chemistry, the <span class="margin_term"><a class="glossterm">concentration</a><span class="glossdef">The quantity of solute that is dissolved in a particular quantity of solvent or solution.</span></span> of a solution describes the quantity of a solute that is contained in a particular quantity of solvent or solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for reactions that occur in solution. Chemists use many different ways to define concentrations, some of which are described in this section.</p>
<div class="section" id="averill_1.0-ch04_s02_s01">
<h2 class="title editable block">Molarity</h2>
<p class="para block" id="averill_1.0-ch04_s02_s01_p01">The most common unit of concentration is <em class="emphasis">molarity</em>, which is also the most useful for calculations involving the stoichiometry of reactions in solution. The <span class="margin_term"><a class="glossterm">molarity (M)</a><span class="glossdef">A common unit of concentration that is the number of moles of solute present in exactly 1 L of solution <span class="inlineequation"><math xml:id="averill_1.0-ch04_m004" display="inline"><semantics><mrow><mrow><mo>(</mo><mrow><mtext>mol/L</mtext></mrow><mo>)</mo></mrow><mo>.</mo></mrow></semantics></math></span></span></span> of a solution is the number of moles of solute present in exactly 1 L of solution. Molarity is also the number of millimoles of solute present in exactly 1 mL of solution:</p>
<div class="equation block" id="averill_1.0-ch04_s02_s01_eq01">
<p class="title editable"><span class="title-prefix">Equation 4.4</span> </p>
<math xml:id="averill_1.0-ch04_m005" display="block">
<semantics>
<mrow>
<mtext>molarity </mtext>
<mo>=</mo>
<mfrac>
<mrow>
<mtext>moles of solute</mtext>
</mrow>
<mrow>
<mtext>liters of solution</mtext>
</mrow>
</mfrac>
<mo>=</mo>
<mfrac>
<mrow>
<mtext>mmoles of solute</mtext>
</mrow>
<mrow>
<mtext>milliliters of solution</mtext>
</mrow>
</mfrac>
</mrow>
</semantics>
</math>
</div>
<p class="para editable block" id="averill_1.0-ch04_s02_s01_p02">The units of molarity are therefore moles per liter of solution (mol/L), abbreviated as M. An aqueous solution that contains 1 mol (342 g) of sucrose in enough water to give a final volume of 1.00 L has a sucrose concentration of 1.00 mol/L or 1.00 M. In chemical notation, square brackets around the name or formula of the solute represent the concentration of a solute. So</p>
<span class="informalequation block">
<span class="mathphrase">[sucrose] = 1.00 M</span>
</span>
<p class="para editable block" id="averill_1.0-ch04_s02_s01_p03">is read as “the concentration of sucrose is 1.00 molar.” The relationships between volume, molarity, and moles may be expressed as either</p>
<div class="equation block" id="averill_1.0-ch04_s02_s01_eq02">
<p class="title editable"><span class="title-prefix">Equation 4.5</span> </p>
<math xml:id="averill_1.0-ch04_m006" display="block">
<semantics>
<mrow>
<msub>
<mi>V</mi>
<mtext>L</mtext>
</msub>
<msub>
<mtext>M</mtext>
<mrow>
<mtext>mol/L</mtext>
</mrow>
</msub>
<mo>=</mo>
<menclose notation="updiagonalstrike">
<mtext>L</mtext>
</menclose>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>mol</mtext>
</mrow>
<mrow>
<menclose notation="updiagonalstrike">
<mtext>L</mtext>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> moles</mtext>
</mrow>
</semantics>
</math>
</div>
<p class="para editable block" id="averill_1.0-ch04_s02_s01_p04">or</p>
<div class="equation block" id="averill_1.0-ch04_s02_s01_eq03">
<p class="title editable"><span class="title-prefix">Equation 4.6</span> </p>
<math xml:id="averill_1.0-ch04_m007" display="block">
<semantics>
<mrow>
<msub>
<mi>V</mi>
<mrow>
<mtext>mL</mtext>
</mrow>
</msub>
<msub>
<mtext>M</mtext>
<mrow>
<mtext>mmol/mL</mtext>
</mrow>
</msub>
<mo>=</mo>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mL</mtext>
</mrow>
</menclose>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>mmol</mtext>
</mrow>
<mrow>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mL</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> mmoles</mtext>
</mrow>
</semantics>
</math>
</div>
<p class="para editable block" id="averill_1.0-ch04_s02_s01_p05">Example 2 illustrates the use of <a class="xref" href="#averill_1.0-ch04_s02_s01_eq02" xrefstyle="select:label">Equation 4.5</a> and <a class="xref" href="#averill_1.0-ch04_s02_s01_eq03" xrefstyle="select:label">Equation 4.6</a>.</p>
<div class="exercises block" id="averill_1.0-ch04_s02_s01_n01">
<h3 class="title">Example 2</h3>
<p class="para" id="averill_1.0-ch04_s02_s01_p06">Calculate the number of moles of sodium hydroxide (NaOH) in 2.50 L of 0.100 M NaOH.</p>
<p class="para" id="averill_1.0-ch04_s02_s01_p07"><strong class="emphasis bold">Given: </strong>identity of solute and volume and molarity of solution</p>
<p class="para" id="averill_1.0-ch04_s02_s01_p08"><strong class="emphasis bold">Asked for: </strong>amount of solute in moles</p>
<p class="para" id="averill_1.0-ch04_s02_s01_p09">
<strong class="emphasis bold">Strategy:</strong>
</p>
<p class="para" id="averill_1.0-ch04_s02_s01_p10">Use either <a class="xref" href="#averill_1.0-ch04_s02_s01_eq02" xrefstyle="select:label">Equation 4.5</a> or <a class="xref" href="#averill_1.0-ch04_s02_s01_eq03" xrefstyle="select:label">Equation 4.6</a>, depending on the units given in the problem.</p>
<p class="para" id="averill_1.0-ch04_s02_s01_p11">
<strong class="emphasis bold">Solution:</strong>
</p>
<p class="para" id="averill_1.0-ch04_s02_s01_p12">Because we are given the volume of the solution in liters and are asked for the number of moles of substance, <a class="xref" href="#averill_1.0-ch04_s02_s01_eq02" xrefstyle="select:label">Equation 4.5</a> is more useful:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch04_m008" display="block">
<semantics>
<mrow>
<mtext>moles NaOH </mtext>
<mo>=</mo>
<msub>
<mi>V</mi>
<mtext>L</mtext>
</msub>
<msub>
<mtext>M</mtext>
<mrow>
<mtext>mol/L</mtext>
</mrow>
</msub>
<mo>=</mo>
<mtext> (2</mtext>
<mtext>.50 </mtext>
<menclose notation="updiagonalstrike">
<mtext>L</mtext>
</menclose>
<mtext>)</mtext>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>0</mtext>
<mtext>.100 mol</mtext>
</mrow>
<mrow>
<menclose notation="updiagonalstrike">
<mtext>L</mtext>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.250 mol NaOH</mtext>
</mrow>
</semantics>
</math>
</span>
<p class="simpara">Exercise</p>
<p class="para" id="averill_1.0-ch04_s02_s01_p13">Calculate the number of millimoles of alanine, a biologically important molecule, in 27.2 mL of 1.53 M alanine.</p>
<p class="para" id="averill_1.0-ch04_s02_s01_p14"><strong class="emphasis bold">Answer: </strong>41.6 mmol</p>
</div>
<p class="para editable block" id="averill_1.0-ch04_s02_s01_p15">Concentrations are often reported on a mass-to-mass (m/m) basis or on a mass-to-volume (m/v) basis, particularly in clinical laboratories and engineering applications. A concentration expressed on an m/m basis is equal to the number of grams of solute per gram of solution; a concentration on an m/v basis is the number of grams of solute per milliliter of solution. Each measurement can be expressed as a percentage by multiplying the ratio by 100; the result is reported as percent m/m or percent m/v. The concentrations of very dilute solutions are often expressed in <em class="emphasis">parts per million</em> (<em class="emphasis">ppm</em>), which is grams of solute per 10<sup class="superscript">6</sup> g of solution, or in <em class="emphasis">parts per billion</em> (<em class="emphasis">ppb</em>), which is grams of solute per 10<sup class="superscript">9</sup> g of solution. For aqueous solutions at 20°C, 1 ppm corresponds to 1 μg per milliliter, and 1 ppb corresponds to 1 ng per milliliter. These concentrations and their units are summarized in <a class="xref" href="#averill_1.0-ch04_s02_s01_t01">Table 4.1 "Common Units of Concentration"</a>.</p>
<div class="table block" id="averill_1.0-ch04_s02_s01_t01" frame="all">
<p class="title"><span class="title-prefix">Table 4.1</span> Common Units of Concentration</p>
<table cellpadding="0" cellspacing="0">
<thead>
<tr>
<th align="center">Concentration</th>
<th align="center">Units</th>
</tr>
</thead>
<tbody>
<tr>
<td>m/m</td>
<td>g of solute/g of solution</td>
</tr>
<tr>
<td>m/v</td>
<td>g of solute/mL of solution</td>
</tr>
<tr>
<td rowspan="2">ppm</td>
<td>g of solute/10<sup class="superscript">6</sup> g of solution</td>
</tr>
<tr>
<td>μg/mL</td>
</tr>
<tr>
<td rowspan="2">ppb</td>
<td>g of solute/10<sup class="superscript">9</sup> g of solution</td>
</tr>
<tr>
<td>ng/mL</td>
</tr>
</tbody>
</table>
</div>
</div>
<div class="section" id="averill_1.0-ch04_s02_s02">
<h2 class="title editable block">The Preparation of Solutions</h2>
<p class="para editable block" id="averill_1.0-ch04_s02_s02_p01">To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution. <a class="xref" href="#averill_1.0-ch04_s02_s02_f01">Figure 4.6 "Preparation of a Solution of Known Concentration Using a Solid Solute"</a> illustrates this procedure for a solution of cobalt(II) chloride dihydrate in ethanol. Note that the volume of the <em class="emphasis">solvent</em> is not specified. Because the solute occupies space in the solution, the volume of the solvent needed is almost always <em class="emphasis">less</em> than the desired volume of solution. For example, if the desired volume were 1.00 L, it would be incorrect to add 1.00 L of water to 342 g of sucrose because that would produce more than 1.00 L of solution. As shown in <a class="xref" href="#averill_1.0-ch04_s02_s02_f02">Figure 4.7 "Preparation of 250 mL of a Solution of (NH"</a>, for some substances this effect can be significant, especially for concentrated solutions.</p>
<div class="figure large editable block" id="averill_1.0-ch04_s02_s02_f01">
<p class="title"><span class="title-prefix">Figure 4.6</span> Preparation of a Solution of Known Concentration Using a Solid Solute</p>
<img src="section_08/421a1cdc4a96997bd44d842d079b4e2e.jpg">
</div>
<div class="figure large editable block" id="averill_1.0-ch04_s02_s02_f02">
<p class="title"><span class="title-prefix">Figure 4.7</span> Preparation of 250 mL of a Solution of (NH<sub class="subscript">4</sub>)<sub class="subscript">2</sub>Cr<sub class="subscript">2</sub>O<sub class="subscript">7</sub> in Water</p>
<img src="section_08/3d3c33730a0c1644e3166fa6be1a8b38.jpg">
<p class="para">The solute occupies space in the solution, so less than 250 mL of water are needed to make 250 mL of solution.</p>
</div>
<div class="exercises block" id="averill_1.0-ch04_s02_s02_n01">
<h3 class="title">Example 3</h3>
<p class="para" id="averill_1.0-ch04_s02_s02_p02">The solution in <a class="xref" href="#averill_1.0-ch04_s02_s02_f01">Figure 4.6 "Preparation of a Solution of Known Concentration Using a Solid Solute"</a> contains 10.0 g of cobalt(II) chloride dihydrate, CoCl<sub class="subscript">2</sub>·2H<sub class="subscript">2</sub>O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of CoCl<sub class="subscript">2</sub>·2H<sub class="subscript">2</sub>O?</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p03"><strong class="emphasis bold">Given: </strong>mass of solute and volume of solution</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p04"><strong class="emphasis bold">Asked for: </strong>concentration (M)</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p05">
<strong class="emphasis bold">Strategy:</strong>
</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p06">To find the number of moles of CoCl<sub class="subscript">2</sub>·2H<sub class="subscript">2</sub>O, divide the mass of the compound by its molar mass. Calculate the molarity of the solution by dividing the number of moles of solute by the volume of the solution in liters.</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p07">
<strong class="emphasis bold">Solution:</strong>
</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p08">The molar mass of CoCl<sub class="subscript">2</sub>·2H<sub class="subscript">2</sub>O is 165.87 g/mol. Therefore,</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch04_m009" display="block">
<semantics>
<mrow>
<msub>
<mrow>
<mtext>moles CoCl</mtext>
</mrow>
<mtext>2</mtext>
</msub>
<msub>
<mrow>
<mtext>•2H</mtext>
</mrow>
<mtext>2</mtext>
</msub>
<mtext>O </mtext>
<mo>=</mo>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>10</mtext>
<mtext>.0 </mtext>
<menclose notation="updiagonalstrike">
<mtext>g</mtext>
</menclose>
</mrow>
<mrow>
<mtext>165</mtext>
<mtext>.87 </mtext>
<menclose notation="updiagonalstrike">
<mtext>g</mtext>
</menclose>
<mtext>/mol</mtext>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.0603 mol</mtext>
</mrow>
</semantics>
</math>
</span>
<p class="para" id="averill_1.0-ch04_s02_s02_p09">The volume of the solution in liters is</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch04_m010" display="block">
<semantics>
<mrow>
<mtext>volume </mtext>
<mo>=</mo>
<mtext> 500 mL</mtext>
<mtext> </mtext>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>1 L</mtext>
</mrow>
<mrow>
<mtext>1000 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mL</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.500 L</mtext>
</mrow>
</semantics>
</math>
</span>
<p class="para" id="averill_1.0-ch04_s02_s02_p10">Molarity is the number of moles of solute per liter of solution, so the molarity of the solution is</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch04_m011" display="block">
<semantics>
<mrow>
<mtext>molarity </mtext>
<mo>=</mo>
<mfrac>
<mrow>
<mtext>0</mtext>
<mtext>.0603 mol</mtext>
</mrow>
<mrow>
<mtext>0</mtext>
<mtext>.500 L</mtext>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.121 M </mtext>
<mo>=</mo>
<msub>
<mrow>
<mtext>CoCl</mtext>
</mrow>
<mtext>2</mtext>
</msub>
<msub>
<mrow>
<mtext>•H</mtext>
</mrow>
<mtext>2</mtext>
</msub>
<mtext>O</mtext>
</mrow>
</semantics>
</math>
</span>
<p class="simpara">Exercise</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p11">The solution shown in <a class="xref" href="#averill_1.0-ch04_s02_s02_f02">Figure 4.7 "Preparation of 250 mL of a Solution of (NH"</a> contains 90.0 g of (NH<sub class="subscript">4</sub>)<sub class="subscript">2</sub>Cr<sub class="subscript">2</sub>O<sub class="subscript">7</sub> in enough water to give a final volume of exactly 250 mL. What is the molar concentration of ammonium dichromate?</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p12"><strong class="emphasis bold">Answer: </strong>(NH<sub class="subscript">4</sub>)<sub class="subscript">2</sub>Cr<sub class="subscript">2</sub>O<sub class="subscript">7</sub> = 1.43 M</p>
</div>
<p class="para editable block" id="averill_1.0-ch04_s02_s02_p13">To prepare a particular volume of a solution that contains a specified concentration of a solute, we first need to calculate the number of moles of solute in the desired volume of solution using the relationship shown in <a class="xref" href="#averill_1.0-ch04_s02_s01_eq02" xrefstyle="select:label">Equation 4.5</a>. We then convert the number of moles of solute to the corresponding mass of solute needed. This procedure is illustrated in Example 4.</p>
<div class="exercises block" id="averill_1.0-ch04_s02_s02_n02">
<h3 class="title">Example 4</h3>
<p class="para" id="averill_1.0-ch04_s02_s02_p14">The so-called D5W solution used for the intravenous replacement of body fluids contains 0.310 M glucose. (D5W is an approximately 5% solution of dextrose [the medical name for glucose] in water.) Calculate the mass of glucose necessary to prepare a 500 mL pouch of D5W. Glucose has a molar mass of 180.16 g/mol.</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p15"><strong class="emphasis bold">Given: </strong>molarity, volume, and molar mass of solute</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p16"><strong class="emphasis bold">Asked for: </strong>mass of solute</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p17">
<strong class="emphasis bold">Strategy:</strong>
</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p18"><strong class="emphasis bold">A</strong> Calculate the number of moles of glucose contained in the specified volume of solution by multiplying the volume of the solution by its molarity.</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p19"><strong class="emphasis bold">B</strong> Obtain the mass of glucose needed by multiplying the number of moles of the compound by its molar mass.</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p20">
<strong class="emphasis bold">Solution:</strong>
</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p21"><strong class="emphasis bold">A</strong> We must first calculate the number of moles of glucose contained in 500 mL of a 0.310 M solution:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch04_m012" display="block">
<semantics>
<mtable>
<mtr>
<mtd>
<msub>
<mi>V</mi>
<mtext>L</mtext>
</msub>
<msub>
<mtext>M</mtext>
<mrow>
<mtext>mol/L</mtext>
</mrow>
</msub>
<mo>=</mo>
<mtext> moles</mtext>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext>500 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mL</mtext>
</mrow>
</menclose>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>1 </mtext>
<menclose notation="updiagonalstrike">
<mtext>L</mtext>
</menclose>
</mrow>
<mrow>
<mtext>1000 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mL</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mtext> </mtext>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>0</mtext>
<mtext>.310 mol glucose</mtext>
</mrow>
<mrow>
<mtext>1</mtext>
<menclose notation="updiagonalstrike">
<mtext>L</mtext>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.155 mol glucose</mtext>
</mtd>
</mtr>
</mtable>
</semantics>
</math>
</span>
<p class="para" id="averill_1.0-ch04_s02_s02_p22"><strong class="emphasis bold">B</strong> We then convert the number of moles of glucose to the required mass of glucose:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch04_m013" display="block">
<semantics>
<mrow>
<mtext>mass of glucose </mtext>
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.155 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mol glucose</mtext>
</mrow>
</menclose>
<mtext> </mtext>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>180</mtext>
<mtext>.16 g glucose</mtext>
</mrow>
<mrow>
<mtext>1 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mol glucose</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> 27</mtext>
<mtext>.9 g glucose</mtext>
</mrow>
</semantics>
</math>
</span>
<p class="simpara">Exercise</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p23">Another solution commonly used for intravenous injections is normal saline, a 0.16 M solution of sodium chloride in water. Calculate the mass of sodium chloride needed to prepare 250 mL of normal saline solution.</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p24"><strong class="emphasis bold">Answer: </strong>2.3 g NaCl</p>
</div>
<p class="para editable block" id="averill_1.0-ch04_s02_s02_p25">A solution of a desired concentration can also be prepared by diluting a small volume of a more concentrated solution with additional solvent. A <span class="margin_term"><a class="glossterm">stock solution</a><span class="glossdef">A commercially prepared solution of known concentration.</span></span>, which is a commercially prepared solution of known concentration, is often used for this purpose. Diluting a stock solution is preferred because the alternative method, weighing out tiny amounts of solute, is difficult to carry out with a high degree of accuracy. Dilution is also used to prepare solutions from substances that are sold as concentrated aqueous solutions, such as strong acids.</p>
<p class="para editable block" id="averill_1.0-ch04_s02_s02_p26">The procedure for preparing a solution of known concentration from a stock solution is shown in <a class="xref" href="#averill_1.0-ch04_s02_s02_f03">Figure 4.8 "Preparation of a Solution of Known Concentration by Diluting a Stock Solution"</a>. It requires calculating the number of moles of solute desired in the final volume of the more dilute solution and then calculating the volume of the stock solution that contains this amount of solute. Remember that diluting a given quantity of stock solution with solvent does <em class="emphasis">not</em> change the number of moles of solute present. The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is therefore</p>
<div class="equation block" id="averill_1.0-ch04_s02_s02_eq01">
<p class="title editable"><span class="title-prefix">Equation 4.7</span> </p>
<span class="mathphrase">(<em class="emphasis">V</em><sub class="subscript">s</sub>)(M<sub class="subscript">s</sub>) = moles of solute = (<em class="emphasis">V</em><sub class="subscript">d</sub>)(M<sub class="subscript">d</sub>)</span>
</div>
<p class="para editable block" id="averill_1.0-ch04_s02_s02_p27">where the subscripts <em class="emphasis">s</em> and <em class="emphasis">d</em> indicate the stock and dilute solutions, respectively. Example 5 demonstrates the calculations involved in diluting a concentrated stock solution.</p>
<div class="figure large editable block" id="averill_1.0-ch04_s02_s02_f03">
<p class="title"><span class="title-prefix">Figure 4.8</span> Preparation of a Solution of Known Concentration by Diluting a Stock Solution</p>
<img src="section_08/d30bd057f60c6bf73ed4f18305fa177c.jpg">
<p class="para">(a) A volume (<em class="emphasis">V</em><sub class="subscript">s</sub>) containing the desired moles of solute (M<sub class="subscript">s</sub>) is measured from a stock solution of known concentration. (b) The measured volume of stock solution is transferred to a second volumetric flask. (c) The measured volume in the second flask is then diluted with solvent up to the volumetric mark [(<em class="emphasis">V</em><sub class="subscript">s</sub>)(M<sub class="subscript">s</sub>) = (<em class="emphasis">V</em><sub class="subscript">d</sub>)(M<sub class="subscript">d</sub>)].</p>
</div>
<div class="exercises block" id="averill_1.0-ch04_s02_s02_n03">
<h3 class="title">Example 5</h3>
<p class="para" id="averill_1.0-ch04_s02_s02_p28">What volume of a 3.00 M glucose stock solution is necessary to prepare 2500 mL of the D5W solution in Example 4?</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p29"><strong class="emphasis bold">Given: </strong>volume and molarity of dilute solution</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p30"><strong class="emphasis bold">Asked for: </strong>volume of stock solution</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p31">
<strong class="emphasis bold">Strategy:</strong>
</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p32"><strong class="emphasis bold">A</strong> Calculate the number of moles of glucose contained in the indicated volume of dilute solution by multiplying the volume of the solution by its molarity.</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p33"><strong class="emphasis bold">B</strong> To determine the volume of stock solution needed, divide the number of moles of glucose by the molarity of the stock solution.</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p34">
<strong class="emphasis bold">Solution:</strong>
</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p35"><strong class="emphasis bold">A</strong> The D5W solution in Example 4 was 0.310 M glucose. We begin by using <a class="xref" href="#averill_1.0-ch04_s02_s02_eq01" xrefstyle="select:label">Equation 4.7</a> to calculate the number of moles of glucose contained in 2500 mL of the solution:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch04_m014" display="block">
<semantics>
<mrow>
<mtext>moles glucose </mtext>
<mo>=</mo>
<mtext> 2500</mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mL</mtext>
</mrow>
</menclose>
<mtext> </mtext>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>1 </mtext>
<menclose notation="updiagonalstrike">
<mtext>L</mtext>
</menclose>
</mrow>
<mrow>
<mtext>1000 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mL</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>0</mtext>
<mtext>.310 mol glucose</mtext>
</mrow>
<mrow>
<mtext>1 </mtext>
<menclose notation="updiagonalstrike">
<mtext>L</mtext>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.775 mol glucose</mtext>
</mrow>
</semantics>
</math>
</span>
<p class="para" id="averill_1.0-ch04_s02_s02_p36"><strong class="emphasis bold">B</strong> We must now determine the volume of the 3.00 M stock solution that contains this amount of glucose:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch04_m015" display="block">
<semantics>
<mrow>
<mtext>volume of stock soln </mtext>
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.775 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mol glucose</mtext>
</mrow>
</menclose>
<mtext> </mtext>
<mrow>
<mo>(</mo>
<mrow>
<mfrac>
<mrow>
<mtext>1 L</mtext>
</mrow>
<mrow>
<mtext>3</mtext>
<mtext>.00 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mol glucose</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
</mrow>
<mo>)</mo>
</mrow>
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.258 L or 258 mL</mtext>
</mrow>
</semantics>
</math>
</span>
<p class="para" id="averill_1.0-ch04_s02_s02_p37">In determining the volume of stock solution that was needed, we had to divide the desired number of moles of glucose by the concentration of the stock solution to obtain the appropriate units. Also, the number of moles of solute in 258 mL of the stock solution is the same as the number of moles in 2500 mL of the more dilute solution; <em class="emphasis">only the amount of solvent has changed</em>. The answer we obtained makes sense: diluting the stock solution about tenfold increases its volume by about a factor of 10 (258 mL → 2500 mL). Consequently, the concentration of the solute must decrease by about a factor of 10, as it does (3.00 M → 0.310 M).</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p38">We could also have solved this problem in a single step by solving <a class="xref" href="#averill_1.0-ch04_s02_s02_eq01" xrefstyle="select:label">Equation 4.7</a> for <em class="emphasis">V</em><sub class="subscript">s</sub> and substituting the appropriate values:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch04_m016" display="block">
<semantics>
<mrow>
<msub>
<mi>V</mi>
<mtext>s</mtext>
</msub>
<mo>=</mo>
<mfrac>
<mrow>
<mtext>(</mtext>
<msub>
<mi>V</mi>
<mtext>d</mtext>
</msub>
<msub>
<mrow>
<mtext>)(M</mtext>
</mrow>
<mtext>d</mtext>
</msub>
<mtext>)</mtext>
</mrow>
<mrow>
<msub>
<mtext>M</mtext>
<mtext>s</mtext>
</msub>
</mrow>
</mfrac>
<mo>=</mo>
<mfrac>
<mrow>
<mtext>(2</mtext>
<mtext>.500 L)(0</mtext>
<mtext>.310 </mtext>
<menclose notation="updiagonalstrike">
<mtext>M</mtext>
</menclose>
<mtext>)</mtext>
</mrow>
<mrow>
<mtext>3</mtext>
<mtext>.00 </mtext>
<menclose notation="updiagonalstrike">
<mtext>M</mtext>
</menclose>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.258 L</mtext>
</mrow>
</semantics>
</math>
</span>
<p class="para" id="averill_1.0-ch04_s02_s02_p39">As we have noted, there is often more than one correct way to solve a problem.</p>
<p class="simpara">Exercise</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p40">What volume of a 5.0 M NaCl stock solution is necessary to prepare 500 mL of normal saline solution (0.16 M NaCl)?</p>
<p class="para" id="averill_1.0-ch04_s02_s02_p41"><strong class="emphasis bold">Answer: </strong>16 mL</p>
</div>
</div>
<div class="section" id="averill_1.0-ch04_s02_s03">
<h2 class="title editable block">Ion Concentrations in Solution</h2>
<p class="para editable block" id="averill_1.0-ch04_s02_s03_p01">In Example 3, you calculated that the concentration of a solution containing 90.00 g of ammonium dichromate in a final volume of 250 mL is 1.43 M. Let’s consider in more detail exactly what that means. Ammonium dichromate is an ionic compound that contains two NH<sub class="subscript">4</sub><sup class="superscript">+</sup> ions and one Cr<sub class="subscript">2</sub>O<sub class="subscript">7</sub><sup class="superscript">2−</sup> ion per formula unit. Like other ionic compounds, it is a strong electrolyte that dissociates in aqueous solution to give hydrated NH<sub class="subscript">4</sub><sup class="superscript">+</sup> and Cr<sub class="subscript">2</sub>O<sub class="subscript">7</sub><sup class="superscript">2−</sup> ions:</p>
<div class="equation block" id="averill_1.0-ch04_s02_s03_eq01">
<p class="title editable"><span class="title-prefix">Equation 4.8</span> </p>
<math xml:id="averill_1.0-ch04_m017" display="block">
<semantics>
<mrow>
<msub>
<mrow>
<msub>
<mrow>
<mtext>(NH</mtext>
</mrow>
<mtext>4</mtext>
</msub>
<mtext>)</mtext>
</mrow>
<mtext>2</mtext>
</msub>
<msub>
<mrow>
<mtext>Cr</mtext>
</mrow>
<mtext>2</mtext>
</msub>
<msub>
<mtext>O</mtext>
<mtext>7</mtext>
</msub>
<mtext>(s)</mtext>
<mover>
<mo>→</mo>
<mrow>
<msub>
<mtext>H</mtext>
<mtext>2</mtext>
</msub>
<mtext>O(l)</mtext>
</mrow>
</mover>
<msub>
<mrow>
<mtext>2NH</mtext>
</mrow>
<mtext>4</mtext>
</msub>
<msup>
<mrow></mrow>
<mo>+</mo>
</msup>
<mtext>(aq)</mtext>
<mo>+</mo>
<msub>
<mrow>
<mtext>Cr</mtext>
</mrow>
<mtext>2</mtext>
</msub>
<msub>
<mtext>O</mtext>
<mtext>7</mtext>
</msub>
<msup>
<mrow>
<msup>
<mrow></mrow>
<mtext>2</mtext>
</msup>
</mrow>
<mo>–</mo>
</msup>
<mtext>(aq)</mtext>
</mrow>
</semantics>
</math>
</div>
<p class="para editable block" id="averill_1.0-ch04_s02_s03_p02">Thus 1 mol of ammonium dichromate formula units dissolves in water to produce 1 mol of Cr<sub class="subscript">2</sub>O<sub class="subscript">7</sub><sup class="superscript">2−</sup> anions and 2 mol of NH<sub class="subscript">4</sub><sup class="superscript">+</sup> cations (see <a class="xref" href="#averill_1.0-ch04_s02_s03_f01">Figure 4.9 "Dissolution of 1 mol of an Ionic Compound"</a>).</p>
<div class="figure large editable block" id="averill_1.0-ch04_s02_s03_f01">
<p class="title"><span class="title-prefix">Figure 4.9</span> Dissolution of 1 mol of an Ionic Compound</p>
<img src="section_08/11367b0b19bd6682f637472684716c36.jpg">
<p class="para">In this case, dissolving 1 mol of (NH<sub class="subscript">4</sub>)<sub class="subscript">2</sub>Cr<sub class="subscript">2</sub>O<sub class="subscript">7</sub> produces a solution that contains 1 mol of Cr<sub class="subscript">2</sub>O<sub class="subscript">7</sub><sup class="superscript">2−</sup> ions and 2 mol of NH<sub class="subscript">4</sub><sup class="superscript">+</sup> ions. (Water molecules are omitted from a molecular view of the solution for clarity.)</p>
</div>
<p class="para editable block" id="averill_1.0-ch04_s02_s03_p03">When we carry out a chemical reaction using a solution of a salt such as ammonium dichromate, we need to know the concentration of each ion present in the solution. If a solution contains 1.43 M (NH<sub class="subscript">4</sub>)<sub class="subscript">2</sub>Cr<sub class="subscript">2</sub>O<sub class="subscript">7</sub>, then the concentration of Cr<sub class="subscript">2</sub>O<sub class="subscript">7</sub><sup class="superscript">2−</sup> must also be 1.43 M because there is one Cr<sub class="subscript">2</sub>O<sub class="subscript">7</sub><sup class="superscript">2−</sup> ion per formula unit. However, there are two NH<sub class="subscript">4</sub><sup class="superscript">+</sup> ions per formula unit, so the concentration of NH<sub class="subscript">4</sub><sup class="superscript">+</sup> ions is 2 × 1.43 M = 2.86 M. Because each formula unit of (NH<sub class="subscript">4</sub>)<sub class="subscript">2</sub>Cr<sub class="subscript">2</sub>O<sub class="subscript">7</sub> produces <em class="emphasis">three</em> ions when dissolved in water (2NH<sub class="subscript">4</sub><sup class="superscript">+</sup> + 1Cr<sub class="subscript">2</sub>O<sub class="subscript">7</sub><sup class="superscript">2−</sup>), the <em class="emphasis">total</em> concentration of ions in the solution is 3 × 1.43 M = 4.29 M.</p>
<div class="exercises block" id="averill_1.0-ch04_s02_s03_n01">
<h3 class="title">Example 6</h3>
<p class="para" id="averill_1.0-ch04_s02_s03_p04">What are the concentrations of all species derived from the solutes in these aqueous solutions?</p>
<ol class="orderedlist" id="averill_1.0-ch04_s02_s03_l01" numeration="loweralpha">
<li>0.21 M NaOH</li>
<li>3.7 M (CH<sub class="subscript">3</sub>)CHOH</li>
<li>0.032 M In(NO<sub class="subscript">3</sub>)<sub class="subscript">3</sub>
</li>
</ol>
<p class="para" id="averill_1.0-ch04_s02_s03_p05"><strong class="emphasis bold">Given: </strong>molarity</p>
<p class="para" id="averill_1.0-ch04_s02_s03_p06"><strong class="emphasis bold">Asked for: </strong>concentrations</p>
<p class="para" id="averill_1.0-ch04_s02_s03_p07">
<strong class="emphasis bold">Strategy:</strong>
</p>
<p class="para" id="averill_1.0-ch04_s02_s03_p08"><strong class="emphasis bold">A</strong> Classify each compound as either a strong electrolyte or a nonelectrolyte.</p>
<p class="para" id="averill_1.0-ch04_s02_s03_p09"><strong class="emphasis bold">B</strong> If the compound is a nonelectrolyte, its concentration is the same as the molarity of the solution. If the compound is a strong electrolyte, determine the number of each ion contained in one formula unit. Find the concentration of each species by multiplying the number of each ion by the molarity of the solution.</p>
<p class="para" id="averill_1.0-ch04_s02_s03_p10">
<strong class="emphasis bold">Solution:</strong>
</p>
<ol class="orderedlist" id="averill_1.0-ch04_s02_s03_l02" numeration="loweralpha">
<li>
<p class="para">Sodium hydroxide is an ionic compound that is a strong electrolyte (and a strong base) in aqueous solution:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch04_m018" display="block">
<semantics>
<mrow>
<mtext>NaOH(s)</mtext>
<mover>
<mo>→</mo>
<mrow>
<msub>
<mtext>H</mtext>
<mtext>2</mtext>
</msub>
<mtext>O(l)</mtext>
</mrow>
</mover>
<msup>
<mrow>
<mtext>Na</mtext>
</mrow>
<mo>+</mo>
</msup>
<mtext>(aq)</mtext>
<mo>+</mo>
<msup>
<mrow>
<mtext>OH</mtext>
</mrow>
<mo>–</mo>
</msup>
<mtext>(aq)</mtext>
</mrow>
</semantics>
</math>
</span>
<p class="para" id="averill_1.0-ch04_s02_s03_p11"><strong class="emphasis bold">B</strong> Because each formula unit of NaOH produces one Na<sup class="superscript">+</sup> ion and one OH<sup class="superscript">−</sup> ion, the concentration of each ion is the same as the concentration of NaOH: [Na<sup class="superscript">+</sup>] = 0.21 M and [OH<sup class="superscript">−</sup>] = 0.21 M.</p>
</li>
<li>
<p class="para"><strong class="emphasis bold">A</strong> The formula (CH<sub class="subscript">3</sub>)<sub class="subscript">2</sub>CHOH represents 2-propanol (isopropyl alcohol) and contains the –OH group, so it is an alcohol. Recall from <a class="xref" href="averill_1.0-ch04_s01#averill_1.0-ch04_s01">Section 4.1 "Aqueous Solutions"</a> that alcohols are covalent compounds that dissolve in water to give solutions of neutral molecules. Thus alcohols are nonelectrolytes.</p>
<p class="para" id="averill_1.0-ch04_s02_s03_p12"><strong class="emphasis bold">B</strong> The only solute species in solution is therefore (CH<sub class="subscript">3</sub>)<sub class="subscript">2</sub>CHOH molecules, so [(CH<sub class="subscript">3</sub>)<sub class="subscript">2</sub>CHOH] = 3.7 M.</p>
</li>
<li>
<p class="para"><strong class="emphasis bold">A</strong> Indium nitrate is an ionic compound that contains In<sup class="superscript">3+</sup> ions and NO<sub class="subscript">3</sub><sup class="superscript">−</sup> ions, so we expect it to behave like a strong electrolyte in aqueous solution:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch04_m019" display="block">
<semantics>
<mrow>
<msub>
<mrow>
<mtext>In(NO</mtext>
</mrow>
<mtext>3</mtext>
</msub>
<msub>
<mtext>)</mtext>
<mtext>3</mtext>
</msub>
<mtext>(s)</mtext>
<mover>
<mo>→</mo>
<mrow>
<msub>
<mtext>H</mtext>
<mtext>2</mtext>
</msub>
<mtext>O(l)</mtext>
</mrow>
</mover>
<msup>
<mrow>
<mtext>In</mtext>
</mrow>
<mrow>
<mtext>3+</mtext>
</mrow>
</msup>
<mtext>(aq)</mtext>
<mo>+</mo>
<msub>
<mrow>
<mtext>3NO</mtext>
</mrow>
<mtext>3</mtext>
</msub>
<msup>
<mrow></mrow>
<mo>–</mo>
</msup>
<mtext>(aq)</mtext>
</mrow>
</semantics>
</math>
</span>
<p class="para" id="averill_1.0-ch04_s02_s03_p13"><strong class="emphasis bold">B</strong> One formula unit of In(NO<sub class="subscript">3</sub>)<sub class="subscript">3</sub> produces one In<sup class="superscript">3+</sup> ion and three NO<sub class="subscript">3</sub><sup class="superscript">−</sup> ions, so a 0.032 M In(NO<sub class="subscript">3</sub>)<sub class="subscript">3</sub> solution contains 0.032 M In<sup class="superscript">3+</sup> and 3 × 0.032 M = 0.096 M NO<sub class="subscript">3</sub><sup class="superscript">–</sup>—that is, [In<sup class="superscript">3+</sup>] = 0.032 M and [NO<sub class="subscript">3</sub><sup class="superscript">−</sup>] = 0.096 M.</p>
</li>
</ol>
<p class="simpara">Exercise</p>
<p class="para" id="averill_1.0-ch04_s02_s03_p14">What are the concentrations of all species derived from the solutes in these aqueous solutions?</p>
<ol class="orderedlist" id="averill_1.0-ch04_s02_s03_l03" numeration="loweralpha">
<li>0.0012 M Ba(OH)<sub class="subscript">2</sub>
</li>
<li>0.17 M Na<sub class="subscript">2</sub>SO<sub class="subscript">4</sub>
</li>
<li>0.50 M (CH<sub class="subscript">3</sub>)<sub class="subscript">2</sub>CO, commonly known as acetone</li>
</ol>
<div class="informalfigure small">
<img src="section_08/0bcbb4d4337db8029b9cebdbcc72a918.jpg">
</div>
<p class="para" id="averill_1.0-ch04_s02_s03_p15">
<strong class="emphasis bold">Answer:</strong>
</p>
<ol class="orderedlist" id="averill_1.0-ch04_s02_s03_l04" numeration="loweralpha">
<li>[Ba<sup class="superscript">2+</sup>] = 0.0012 M; [OH<sup class="superscript">−</sup>] = 0.0024 M</li>
<li>[Na<sup class="superscript">+</sup>] = 0.34 M; [SO<sub class="subscript">4</sub><sup class="superscript">2−</sup>] = 0.17 M</li>
<li>[(CH<sub class="subscript">3</sub>)<sub class="subscript">2</sub>CO] = 0.50 M</li>
</ol>
</div>
<div class="key_takeaways block" id="averill_1.0-ch04_s02_s03_n02">
<h3 class="title">Key Equations</h3>
<p class="para">
<strong class="emphasis bold">definition of molarity</strong>
</p>
<p class="para"><a class="xref" href="#averill_1.0-ch04_s02_s01_eq01" xrefstyle="select:label">Equation 4.4</a>: <span class="inlineequation"><math xml:id="averill_1.0-ch04_m020" display="inline"><semantics><mrow><mtext>molarity </mtext><mo>=</mo><mfrac><mrow><mtext>moles of solute</mtext></mrow><mrow><mtext>liters of solution</mtext></mrow></mfrac><mo>=</mo><mfrac><mrow><mtext>mmoles of solute</mtext></mrow><mrow><mtext>milliliters of solution</mtext></mrow></mfrac></mrow></semantics></math></span></p>
<p class="para">
<strong class="emphasis bold">relationship among volume, molarity, and moles</strong>
</p>
<p class="para"><a class="xref" href="#averill_1.0-ch04_s02_s01_eq02" xrefstyle="select:label">Equation 4.5</a>: <span class="inlineequation"><math xml:id="averill_1.0-ch04_m021" display="inline"><semantics><mrow><msub><mi>V</mi><mtext>L</mtext></msub><msub><mtext>M</mtext><mrow><mtext>mol/L</mtext></mrow></msub><mo>=</mo><menclose notation="updiagonalstrike"><mtext>L</mtext></menclose><mtext> </mtext><mrow><mo>(</mo><mrow><mfrac><mrow><mtext>mol</mtext></mrow><mrow><menclose notation="updiagonalstrike"><mtext>L</mtext></menclose></mrow></mfrac></mrow><mo>)</mo></mrow><mo>=</mo><mtext> moles</mtext></mrow></semantics></math></span></p>
<p class="para">
<strong class="emphasis bold">relationship between volume and concentration of stock and dilute solutions</strong>