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Analytical Chemistry Homework
Due Wednesday Week 6
Problems 6-26: 9-17,23,29: 7-7,8,11
6-26. A solution contains 0.0500 M Ca2+ and 0.0300 M Ag+. Can 99.00% of either ion
be precipitated by addition of sulfate without precipitating the other metal ion? What will
be the concentration of Ca2+ when Ag2SO4 begins to precipicate?
Physical – a solution with [Ca2+] = 0.0500 M and [Ag+] = 0.0300 M. “Can 99% of either
ion be precipitated by addition of sulfate, without precipitating the other metal ion?” We
are going to add sulfate until one precipitates until only 1% of it is left as ions in solution.
Which one?
Ksp for silver sulfate is
1.5 x 10-5
Ksp for calcium sulfate is 2.4 x 10-5
Ksp for barium sulfate is 1.1 x 10-10
Ksp = [Ca2+][SO42-]
Ag2SO4(s) <= => 2 Ag+ + SO42CaSO4(s) <= => Ca2+ + SO42BaSO4(s) <= => Ba2+ + SO42-
Ksp = [Ag+]2 [SO42-]
Calcium least soluble – will precipitate first. Final [Ca2+] = .000500 M (1% of initial)
At this point the equilibrium with sulfate will be
[SO42-] = Ksp / [Ca2+] = 2.4 x 10-5/0.000500 = 0.048 M
__________
____________
+
2So [Ag ] =  Ksp/ [SO4 ] =  1.5x10-5/0.048 = .0177 M
this is lower than .0300 M so some of the silver has already precipitated at this level of
sulfate.
.0177/0.0300 x 100% = 59% so 59% of the original silver is left in solution and 41% has
precipitated.
What can we learn from this? We can not use sulfate precipitation to remove a calcium
ion interferent from a silver solution. Nor can we analyze calcium ion with this methos
in the presence of silver ion.
What about the second half of the question? When does silver sulfate begin to
precipitate?
[SO42-] = Ksp/ [Ag+]2
= 1.5 x 10-5/ (0.0300)2 = 0.01667 M
[Ca2+] = Ksp /[SO42-] = 2.4 x 10-5 / 0.01667 = 1.4 x 10-3 M
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1.4 x 10-3 M/0.0500 M x 100% = 2.8%
When silver sulfate starts to precipitate, 97% of the calcium has precipitated. And when
calcium from 97 to 99% precipitated, silver ion goes from 0 to 41% precipitated.
9-17 a) Why do many rivers in Box 9-1. lie on the line [HCO3-] = 2[Ca2+]?
According to Box 9-1, the source of calcium in the rivers is the mineral calcite, which
dissolves by reacting with carbon dioxide in the river waver according to the equation:
CaCO3(s) + CO2(aq) + H2O < == > Ca2+ + 2 HCO3If the predominate product is bicarbonate and not carbonate or carbonic acid,then the
mass balance of this reaction is [HCO3-] = 2[Ca2+]. Rivers on the line [HCO3-] = 2[Ca2+]
are saturated with calcite.
b) What is happening in rivers that lie above the line [HCO3-] = 2[Ca2+]?
Rivers above the line [HCO3-] = 2[Ca2+] contain more bicarbonate than Ca2+. These
rivers are not saturated with calcite.
c) The Rio Grande lies below the line [HCO3-] = 2[Ca2+]. Propose a hypothesis for what
might be happening in this river.
The Rio Grande has more Ca2+ than expected from dissolved calcite. Therefore there is
probably another source of calcium ions than calcite. This source is likely to be more
soluble than calcite. For example, CaSO4 found in anhydrite and gypsum is 104 times
more soluble than calcite. Perhaps the Rio Grande flows over minerals like this that are
more soluble than calcite.
9-23. a) Charge balance
[NH4+] + [H+] = 2 [SO42-] + [HSO4-] + [OH-]
b)
mass balance
[NH3] + [NH4+] = 2 [SO42-] + 2 [HSO4-]
c) Want to determine ammonia concentration so we will aim to get everything solved for
ammonia.
[H+] = 10-9.25 M = 5.62 x 10-10 so [OH-] = 1.00 x 10-14 / 5.62 x 10-10 = 1.78 x 10-5
Ka = [NH3][H+]/[NH4+] solving for ammonium ion gives
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[NH4+] = [NH3][H+]/ Ka = 5.62 x 10-10/ 5.70 x 10-10 [NH3] = 0.986 [NH3]
Kb = [HSO4-][OH-]/[SO42-] solving for hydrogen sulfate gives
[HSO4-] = Kb [SO42-]/[OH-] = 9.80 x 10-13/1.78 x 10-5 [SO42-] = 5.51x10-8 [SO42-]
now let’s substitute what we know into the mass balance equation
[NH3] + [NH4+] = 2 [SO42-] + 2 [HSO4-]
[NH3] + 0.986 [NH3] = 2 [SO42-] + 2 x 5.51x10-8 [SO42-] = 2 + 1.10x10-7 [SO42-]
1.986 [NH3] = 2 [SO42-]
 [SO42-] = 0.993 [NH3]
Now putting all of the ammonia terms into the Ksp gives
Ksp = [NH4+]2[SO42-] = {0.986 [NH3]}2 {0.993[NH3]} = 0.9654 [NH3]3
[NH3]3 = 276/0.9654 = 286
[NH3] = 6.59 M
7-7. How many milliliters of 0.100 M KI are needed to react with 40.0 mL of 0.0400 M
Hg2(NO3)2 if the reaction is Hg22+ + 2 I-  Hg2I2(s)?
40.0 mL x 0.0400 mmol Hg22+/mL x 2 mmol I-/1 mmol Hg22+ x 1 mL/0.100 mmol I- =
32.0 mL
7-8. For reaction 7-1, how many milliliters of 0.1650 M KMnO4 are needed to react with
108.0 mL of 0.1650 oxalic acid? How many milliliters of 0.1650 M oxalic acid are
required to reaction with 108.0 mL of 0.1650 M KMnO4?
The chemical equation is (using C2O4H2 for oxalic acid)
5 C2O4H2 + 2 MnO4- + 6 H+  10 CO2 + 2 Mn2+ + 8 H2O
The long way to do this is through the stoichiometry
108.0 mL x 0.1650 mmol oxalic/mL x 2 mol MnO4-/5 mol oxalic x 1 mL/0.1650 mol MnO4= 43.20 mL MnO4- solution
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An easy way is to note that the reagents have the same concentration. Therefore, the
volume of permanganate needed is 2/5 of the volume of oxalic acid.
i.e. 108.0 mL x 2/5 = 43.0 mL
For the second part of the question
Volume of oxalic acid needed is = 5/2 volume of MnO4- = 108.0 x 5/2 = 270.0 mL
7-11. Limestone consists mainly of the mineral calcite, CaCO3. The carbonate content
of 0.5413g of powdered limestone was measured by suspending the powder in water,
adding 10.00 mL of 1.396 M HCl, and heating to dissolve the solid and expel CO2.
CaCO3(s) + 2 H+ == > Ca2+ + CO2(g) + H2O
FW 100.087
The excess acid required 39.96 mL of 0.1004 M NaOH for complete titration to a
phenolphthalein end point. Find the weight percent of calcite in the limestone.
This is like a back titration. Some of the acid reacts with carbonate in the above reaction
and the rest reacts with the hydroxide. We get the amount that reacts with carbonate by
knowing the total moles of acid and subtracting the amount that reacts with hydroxide in
the titration.
Total HCl is 10.00 mL x 1.396 mmol/mL = 13.96 mmol H+
The hydroxide required is 39.96 mL x 0.1004 mmol/mL = 4.012 mmol OHSince 1 OH- reacts with one H+, the titration reacted with 4.012 mmol H+
Therefore, 13.96 – 4.012 = 9.948 mmol H+ were consumed in the reaction with
carbonate.
9.948 mmol H+ x 1 mmol CaCO3/2 mmol H+ x 100.087 mg/mmol = 497.8 mg of CaCO3
Weight Percent in sample is
0.4978 CaCO3/ 0.5413 x 100% = 91.96 % = 92.0 wt %