1
Analytical Chemistry
Solutions to Workshop Problems due Week 2
1.
(a) Calculate the p-function of the ion-product constant of water, Kw, at 25C.
(b) Calculate the p-function of the concentrations of the four ions in a solution that is
simultaneously 2.00 x 10-3 M in NaCl and 5.4 x 10-4 M in HCl. [Look at part (a) to get a hint as
to the nature of the fourth ion.]
(a) The value of the ion product of water, Kw is given on page 101 and a Table 6-1 shows its
temperature dependence. The value is 1.01 x 10-14 at 25C
pKw = - log (1.01 x 10-14) = 13.996 (a value of 14.0 or 14.00 is fine)
(b)
The concentration of H+ is equal to the concentration of HCl since one H+ is present in solution for
each HCl molecule that was dissolved in the solution. The same is true of the Na+ cation. It’s
concentration is equal to the concentration of the NaCl in the solutions since one Na+ is present fro
each NaCl molecule that dissolved.
So
pH = - log [H+] = - log (5.4 x 10-4) = -(-3.2676) = 3.27
pNa = - log (2.00 x 10-3) = 2.699
Notice that these answers have different significant figures as the given concentrations have different
sig figs.
Cloride anions in the solution come from both the HCl and NaCl so the total Cl- concentration is given
by the sum of the concentrations of the two solutes:
[Cl-] = 2.00 x 10-3 M + 5.4 x 10-4 M = 2.54 x 10-3 M
pC1 = - (log 2.54 x 10-3) = 2.595
Finally the other anion present is the hydroxide ion. If you know the concentration of H+ in a solution
you can use Kw to calculate the hydroxide ion concentration.
Kw = [H+][OH-]
solving for hydroxide gives
[OH-] = Kw/ [H+]
[OH-] = Kw/ [H+] = 1.00 x 10-14 / 5.4 x 10-4 = 1.852 x 10-11
pOH = - log (1.852 x 10-11 ) = 10.73
(has 2 sig figs since 5.4 has two sig figs)
If you understand logs well you remember that if dividing numbers is the same as subtracting their
logs. So a quick way to get pOH if pH is know is the following
pOH = 14.00 – pH = 14.00 - 3.27 = 10.73
2
2. Magnetite is a mineral having the formula Fe3O4 or FeO·Fe2O3. A 1.1324-g sample of a
magnetite ore was dissolved in concentrated HCl to give a solution that contained a mixture of
Fe2+ and Fe3+ ions. Nitric acid was added and the solution was boiled for a few minutes, which
converted all of the iron ions to Fe3+. The Fe3+ was then precipitated as Fe2O3·xH2O by addition
of NH3. {The x here refers to a variable amount of water being included in the solid precipitate.]
After filtration and washing, the residue was ignited at a high temperature to give 0.5394 g of
pure Fe2O3 (159.69 g/mol). [Heating drives off the water so that a stoichiometric solid with the
formula Fe2O3 is left behind.] Calculate (a) the percent by weight of Fe (55.847 g/mol) in the
original ore sample and (b) the percent Fe3O4 (231.54 g/mol) in the original ore sample.
Like most stoichiometry problems you start with the known information (usually given at the end of
the problem) and convert the known information into moles, use a stoichiometric ratio between the
known and the unknown substances and then calculate the decided property of the later.
Here we know that we have 0.5394 g of Fe2O3 at the end of the assay. We assume that all the iron in
the original sample is in this solid and want to know the mass of just the iron in the sample. If we then
divide the mass of iron by the mass of the sample, we can determine the percent of iron in the sample.
As we will do and discuss over and over again, we take the mass of one substance, convert to moles of
that substance using its atomic or molecular mass, multiply by a molar ratio between the two
substances and convert the moles of the second substance into the desired property of the second
substance. Follow these steps in the equation below
(a) mass Fe = 0.5394 g Fe2O3 x
1 mol Fe2O3 x 2 mol Fe x 55.847 g Fe
159.69 g Fe2O3 1 mol Fe2O3
1 mol Fe
= 0.37728 g Fe
The first step converts to moles of Fe2O3. The next step uses the fact that there are 2 mol of iron in
each mole of Fe2O3. Now we have moles of Fe. Multiplying by the atomic mass gives mass of
iron in the original sample.
We now get percent iron in the original sample by dividing by the mass of the sample and turning
this ratio into a percent by multiplying by 100%.
percent Fe = 0.37728 g
1. 1324 g
x 100% = 33.317 = 33.32%
How many sig figs? We did our weighing to 4, so the answer has 4 sig figs. Why doe the 1’s and
2’s in the equation not affect the number of sig figs?
(b) This part is fairly difficult. Like the problem above for iron, we need to know the mass of
Fe3O4 in the sample in order to determine is percent. There are several ways to do this. Below
are two of these. (1) From the formula Fe3O4 we see that there are 3 iron atoms for every Fe3O4
formula mass. We see that we found the number of moles of iron half way through the above
calculation. So we can at that point convert the moles Fe to moles of Fe3O4 and then find the
mass of Fe3O4.
3
(1) mass Fe3O4 = 0.5394 g Fe2O3 x
1 mol Fe2O3 x 2 mol Fe x 1 mole Fe3O4 x 231.54 g Fe3O4
159.69 g Fe2O3 1 mol Fe2O3
3 mol Fe
1 mole Fe3O4
= 0.521396 g Fe3O4
percent Fe3O4 = 0.521396 g Fe3O4 x 100% = 46.043 = 46.04%
1.324 g
Still 4 sig figs.
(2) In this approach, we find a molar ratio between Fe2O3 in the final solid and the Fe3O4 in the
original sample. The chemical equation below shows that there are 3 mol of Fe2O3 formed
from every 2 mol of Fe3O4 present originally.
3 Fe2O3
== > 2 Fe3O4 + ½ O2
Now our equation is shorter, but still has the same answer.
mass Fe3O4 = 0.5394 g Fe2O3 x 1 mol Fe2O3 x 2 mol Fe3O4 x 231.54 g Fe3O4
159.69 g Fe2O3
3 mol Fe2O3
1 mol Fe3O4
= 0.521396 g
Problems from Harris:
6-5 For the reaction 2 A(g) + B(aq) + 3 C(l) < === > D(s) + 3 E(g), the concentrations at
equilibrium are found to be
A:
B:
C:
D:
E:
2.8 x 103 Pa x 1 bar / 105 Pa = 0.028 bar
1.2 x 10-2 M
12.8 M
16.5 M
3.6 x 104 Torr x 133.322 Pa/1 Torr x 1 bar/ 105 Pa = 47.996 bar
Find the numerical value of the equilibrium constant that would appear in a conventional table
of equilibrium constants.
First we need to remember that pure liquids and solids do not appear in equilibrium calculations. So
substances C and D are not used in the calculation. Next we have two gases A and E that need to
have their partial pressures converted to bars. I did this above using the conversion factors found in
Table 1-4 on page 13.
Now we put products over reactants to get an equilibrium expression
K=
P E3
. =
(47.996)3
PA2 [B]
(0.028)2 (1.2 x 10-2)
= 1.2 x 1010
6-6 From the reactions
HOCl < === > H+ + OCl-
K = 3.0 x 10-8
(2 sig figs)
4
HOCl + OBr- < === > HOBr + OCl-
K = 15
Find the value of K for the reaction HOBr < === > H+ + OBr- . All species are aqueous.
[The methods for this problem are found on pps. 100-101.]
We want to combine these two reactions in such a way that the “make” the desired reaction. We see
first that HOBr is a reactant in the desired reaction. In order to get it on the left side we will have to
reverse the second reaction like this
HOBr + OCl- < === > HOCl + OBr-
K = 1/15
When we reverse a reaction we get the inverse of the K, so K for this new reaction is as shown above.
If we add this reversed reaction with the first reaction above, we will be able to eliminate all the
chloride containing species.
HOBr + OCl- < === > HOC l + OBrHOCl < === > H+ + OCl-
K = 1/15
K = 3.0 x 10-8
This leaves us with the desired chemical equation and the rule for adding reactions is that we multiply
their Ks, so that
HOBr < === > H+ + OBr-
K = 1/15 x 3.0 x 10-8 = 2.0 x 10-9
6-8
For the reaction HCO3- < === > H+ + CO32-, ∆G = + 59.0 kJ/mol at 298.15 K. Find
the value of K for the reaction.
Page 102 (and lecture) give a relationship between the equilibrium constant K and the free energy
of a reaction ∆G. We use the formula (and the example on page 102)
K = e-∆G/RT = e-(59.0 kJ/mol x 1000 J/kJ)/(8.314472 J/mol/K)(298.15 K) = 4.61 x 10-11
A few words about this equation. We need to convert the value of ∆G from kJ to J so that the units
will cancel the J in the gas constant R. Values of constants are on the inside back cover of the
book. The Temperature always has the units of degrees K, so if it is given in other units you will
need to convert. This time it was already in K.
6-12 For the reaction H2(g) + Br2(g) < === > 2 HBr(g), K = 7.2 x 10-4 at 1362K and ∆Ho is
positive. A vessel is charged with 48.0 Pa HBr, 1370 Pa of H2, and 3310 Pa Br2 at 1362 K.
(a) Will the reaction proceed to the left or right to reach equilibrium?
We use the reaction quotient Q the determine if a system is at equilibrium and the direction in
which equilibrium lies. [from page 103] For this reaction
Q = PHBr2 .
PH2∙ PBr2
5
Let’s convert the partial pressures to bars first.
HBr: 48.0 Pa x 1 bar/ 105 Pa = 4.8 x 10-4 bar
H2 : 1370 Pa x 1 bar/ 105 Pa = 0.01370 bar
Br2: 3310 Pa x 1 bar/ 105 Pa = .03310 bar
Q = PHBr2 .
PH2∙ PBr2
=
(4.8 x 10-4)2 = 5.1 x 10-4 this is less than 7.2 x 10-4 so we need to
0.01370 ∙ 0.03310
Increase the numerator to get to K so the reaction goes to the right (forms more product).
(b) Calculate the pressure in Pa of each species in the vessel at equilibrium.
Set up a concentration table
H2(g) + Br2(g) < === > 2 HBr(g)
Initial pressure
Eq. pressure
0.0137 0.0331
0.0137 – x 0.0331 – x
4.8 x 10-4
4.8 x 10-4 + 2x
Remember that 2x of HBr are formed when x of H2 is consumed.
K = 7.2 x 10-4 = (4.8 x 10-4 + 2x)2 this looks like messy algebra. Let’s assume that the x is
(.0137-x)(.0331-x) small compared to the reactant values.
This gives us
(4.8 x 10-4 + 2x)2 = 7.2 x 10-4 (.0137)(0.0331) = 3.265 x 10-7
take the square root of both sides
4.8 x 10-4 + 2x = 5.714 x 10-4
2x = 9.140 x 10-5
x = 4.50 x 10-5 bar x 105 Pa/bar = 4.50 Pa (this is 0.0000450 bar which is small compared to
.01370 and .03310 so our assumption was acceptable)
So at Equilibrium we have
HBr: 48.0 Pa + 2 x 4.5 = 57.0 Pa
H2 : 1370 – 4.5 = 1366 Pa
Br2: 3310 - 4.5 = 3306 Pa
c) If the mixture at equilibrium is compressed to half its original volume, will the reaction
proceed to the left or the right to reestablish equilibrium?
If we decrease the volume to ½ its original the pressures will double. [This is the old ideal gas law.
As long as temperature or moles do not change then PV = nRT says that PV is a constant. So if V
goes down by ½, P must go up by a factor of 2 for PV to remain constant.
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What does this do to the K?
K = PHBr2 .
PH2∙ PBr2
The pressure of each component doubles, but this means 22 change in the numerator and a 2 x 2
change in the denominator. So although the pressures all double, the K value does not change.
d) If the mixture at equilibrium is heated from 1362 to 1407 K, will HBr be formed or
consumed in order to reestablish equilibrium?
See the bottom of page 103 and top of 104. The reaction is endothermic (∆H is positive) so K will
increase as the temperature is increased. A larger value of K means the numerator must go up so
product concentration is increased, i.e. we form HBr.
Another way to look at this is that an endothermic reaction absorbs heat when it goes from reactants
to products.
Heat + reactants < === > products.
Le Chatelier’s principle says that if we add heat the reaction will go in the direction to reduce the
amount of added heat. This is in the direction of forming products.
6-14 Use the solubility product to calculate the solubility of CuBr (FM 143.45) in water,
expressed as (a) moles per liter; (b) grams per 100 mL.
(a) Solubility products are written as the dissociation
CuBr(s) < === > Cu1+ + Br1Initial conc.
Final conc.
solid
solid
0
x
0
x
Ksp = [Cu2+][Br-]
Solubility products are in Appendix F. The value of Ksp for copper(II)
bromide is given as 5 x 10-9.
x2 = 5 x 10-9

x = √ (5 x 10-9) = 7.1 x 10-5 M (really only 1 sig fig)
(b) So we have moles per liter and need to change to grams per 100 mL, so this is a unit coversion
step.
7.1 x 10-5 mol/L x 143.45 g/mol x 0.1 L / 100 mL = 1.0 x 10-3 g/100 mL (again only 1 sig fig)
6-15 Use the solubility product to calculate the solubility of
Ag4Fe(CN)6 (FM 643.42) < === > 4 Ag+ + Fe(CN)64- in water expressed as (a) moles
per liter; (b) grams per 100 mL; (c) ppb Ag+ (~ ng Ag+ /mL).
7
Ag4Fe(CN)6 < === > 4 Ag+ + Fe(CN)64Initial
Final
solid
solid
0
4x
0
x
(a)
Ksp = 8.5 x 10-45 = (4x)4 ∙ x = 256x5 (value from Appendix F)
x =
(8.5 x 10-45/256)1/5 = 5.06 x 10-10 M (two sig figs but carrying an extra for the next part)
Since one mole of ferrocyanide ion is released for each mole of silver ferrocyanide that dissolves,
the value of x is the solubility of silver ferrocyanide in moles per liter.
(b) 5.06 x 10-10 mol/L x 643.32 g/mol x 0.1 L/100 mL = 3.3 x 10-8 g/100 mL
c) [Ag+] = 4x = 4(5.06 x 10-10 mol/L) x 107.868 g/mol x 1 L/1000 mL x 1g/109 ng = 0.22 ppb Ag+
Notice here that we have to use the formula mass of silver ferrocyanide in part (b) because we want
g of silver ferrocyande per 100 mL. But in part c) we use the atomic mass of silver because we are
interested in the ppb of silver ion in the solution.
6-19 Find the solubility in g/L of CaSO4 (FM 136.14) in (a) distilled water; (b) 0.50 M CaCl2.
You should be able, by now to write out the chemical equation of the solubility of calcium
sulfate. Remember that solubilities are written as dissociations into ions.
a)
CaSO4(s) < === > Ca2+ + SO42-
Initial
Final
solid
solid
0
x
0
x
Ksp = 2.4 x 10-5 ( from Appendix F)
x2 = 2.4 x 10-5 
x = √ (2.4 x 10-5) = 4.899 x 10-3 M
now get M into g/L
4.899 x 10-3 mol/L x 136.14 g/mol = 0.667 g/L
b) This is a common ion problem. We already have some calcium in the solution before the
calcium sulfate dissolves. From our lecture discussions you should expect that the solubility will
be less in the presence of the additional calcium ion.
CaSO4(s) < === > Ca2+ + SO42Initial
Final
solid
solid
0.50
0.50+ x
0
x
8
x (0.50 + x) = 2.4 x 10-5
we expect that x is less than 5 x 10-3 from part a. Let’s assume that
x is small compared to 0.50 M. Then the equation becomes.
0.50x = 2.4 x 10-5

x = 4.8 x 10-5 M (yes, x is small compared to 0.50)
again convert M to g/L
= 4.8 x 10-5 M x 136.14 g/mol = 6.5 x 10-3 g/L
The solubility has gone down by over a factor of 100!
6-25 If a solution containing 0.10 M Cl-, Br-, I-, and CrO42- is treated with Ag+, in what
order will the anions precipitate?
This is like a common ion problem in that one ion of a silver compound is already in solution.
As we add the silver cation we get to a point where the Ksp for each compound is at equilibrium
and then gets exceeded. When Q exceeds the Ksp the substance will begin to precipitate. The
silver compounds formed are AgCl, AgBr, AgI and Ag2CO4. We want to know the [Ag+] at
equilibrium with 0.1 M of anion. For three of these compounds the Ksp has the form
Ksp = [Ag+][X-] where X- is one of the three halides. For silver chromate Ksp =[Ag+]2[CrO42-]
And the math will be slightly different. =[Ag+]2 = Ksp/[CrO42-]  [Ag+] = √ ( Ksp/[CrO42-] )
We start by looking up the Ksp’s
Salt
Ksp
[Ag+]
AgCl
1.8 x 10-10
Ksp/0.10
=
1.8 x 10-9
AgBr
5.0 x 10-13
Ksp/0.10
=
5.0 x 1013
AgI
8.3 x 10-17
Ksp/0.10
=
8.3 x 10-17
Ag2CO4
1.2 x 10-12
√ (Ksp/0.10
=
3.5 x 10-6
So we see that as the silver ion concentration gets larger, first iodide then bromide, then chloride
and finally chromate will precipitate. If one can collect the precipitate in between these steps, you
can see how one might separate and analyze a mixture of these anion by using silver addition.