CHEM 301 - Assignment #3
SOLUTIONS
1. A water sample is known to have a pH of 6.44 and a bicarbonate ion concentration of 42.4 mg/L
as C. Calculate the total inorganic carbon and report [CO32-]T.
Solution:
The total inorganic carbon is given by;
[CO32-]T = [CO2(aq)] + [HCO3-] + [CO32-]
but inspection of the pH speciation diagram (Fig 1.2, text) or the Ka values, indicates that at
pH = 6.44, [CO32-] << [CO2(aq)] and [HCO3-]
Converting the concentration units;
[HCO3-] = 42.4 mg C/L x 1mol C/12,000 mg x 1mol HCO3-/1 mol C
= 3.53 x 10-3 M
Using this value in the Ka1 expression allows us to calculate the [CO2(aq)] as follows;
−
[ H + ][ HCO3 ]
K a1 =
[CO2 (aq)]
−
[H + ][HCO3 ] (10−6.44 )(3.53 x 10−3 )
=
= 2.85 x 10− 3 M
∴ [CO2 (aq)] =
−7
K a1
4.5 x 10
Thus;
[CO32-]T = 2.85 x 10-3 M + 3.53 x 10-3 M = 6.38 x 10-3 M
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2. Sulfur dioxide dissolves in water and dissociates into hydrogen sulfite. Write these chemical
equilibria and calculate the pH of rainwater equilibrated with 70. ppbv of an SO2 contaminated
atmosphere.
Solution:
We will need to consider both the dissolution of SO2(g) into water (KH) and the subsequent
dissociation to produce protons (Ka1 and Ka2). These equilibria are summarized below.
SO2(g)
====
SO2(aq) + H2O
HSO3- ====
KH = 1.2 x 10-5 M Pa-1 (text, Table 11.1)
SO2(aq)
====
H+ +
H+ + HSO3-
SO32-
Ka1 = 1.72 x 10-2 (text, App. B1)
Ka2 = 6.43 x 10-8 (text, App. B1)
Assuming a total pressure of one atmosphere, the PSO2 = 70 x 10-9 atm.
Converting the given KH into atmospheres yields;
1.2 x 10-5 M Pa-1 x 101,300 Pa atm-1 = 1.22 M atm-1
Thus, [SO2(aq)] = KH x PSO2 = (1.22 M atm-1) (70 x 10-9 atm) = 8.5 x 10-8 M
Using the Ka1 expression and assuming no further dissociation of HSO3- (i.e., ignoring Ka2),
yields;
x2
K a1 =
,
[ SO2 (aq )]
where x = [H+] = [HSO3-] and we assume that the atmosphere provides an inexhaustible
supply of SO2(aq)
∴ [H+]2 = (1.72 x 10-2) (8.5 x 10-8)
so
[H+] = 3.83 x 10-5 M
∴pH = 4.42
Note 1: A much lower pH results from a smaller amount of SO2 than for CO2. This is the
result of two factors KH(SO2) is larger and Ka1(SO2) is larger.
Note 2: If we consider the second dissociation using the Ka2 expression, we solve for an
additional [H+] = (Ka2 x [HSO3-])1/2 = 1.5 x 10-6 M, which makes a relatively insignificant
contribution to the pH when added to the 3.8 x 10-5 M already present from the first
dissociation, [H+] ~ 3.83 x 10-5 + 1.5 x 10-6 = 3.98 x 10-5 M, so pH = 4.40.
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3. Methane and carbon dioxide are produced under anaerobic conditions by the fermentation of
organic matter, approximated by the following equation
2 {CH2O} Æ CH4 + CO2
Gas bubbles are evolved at 5m depth and remain in contact with water at the sediment surface long
enough so that equilibrium is attained. The total pressure at this depth is 148 kPa. Calculate the
aqueous molar concentration of methane in the interstitial water at 25oC. Assuming that the dissolved
carbon dioxide is the only source of acidity, calculate the pH of this water and the total carbonate
concentration.
Solution:
In order to solve this problem, we need find the partial pressures of the individual gases CH4
and CO2 which are in equilibrium with the water. Since this reaction is occurring under
anaerobic conditions, these are the only gases likely to be presence at appreciable
concentrations. Furthermore, since they are produced with 1:1 stoichiometry, it is fair to
assume that the total pressure is given by;
PT = PCO2 + PCH4
and PCO2 = PCH4 = 74 kPa
From the Henry’s Law expression;
[CH4(aq)] = KH PCH4
Using the KH value in Table 11.1 (textbook) yields;
[CH4] = 1.3 x 10-7 M Pa-1 x 74,000 Pa = 9.6 x 10-4 M
To calculate the pH and the total carbonate, we will first need the [CO2].
Proceeding as above for methane;
[CO2] = 3.3 x 10-7 M Pa-1 x 74,000 Pa = 2.44 x 10-2 M
Assuming this to be the only source of [H+] and substituting into the Ka1 expression yields;
x2
K a1 =
[CO2 (aq )]
where x represents [H+] = [HCO3-]
Thus, [H+] = 1.05 x 10-4 M and pH = 4.0
The total carbonate concentration is given by;
[CO32-]T = [CO2(aq)] + [HCO3-] + [CO32-]
at pH 4.0, [CO32-] << [CO2(aq)] and [HCO3-]
so,
[CO32-]T = [CO2(aq)] + [HCO3-]
Since we have already calculated the [HCO3-] from the Ka1 expression (x = 1.05 x 10-4 M)
and the [CO2(aq)] = 2.44 x 10-2 M, the total carbonate is given by;
[CO32-]T = 2.44 x 10-2 M + 1.05 x 10-4 M = 2.45 x 10-2 M ~ 2.5 x 10-2 M
solutions assign 3 2006
4. Use the following chemical equilibria for a water sample in equilibrium with atmospheric CO2(g)
and limestone (CaCO3). In such a system, the concentrations of H+ and Ca2+are coupled to each other
and to the partial pressure of CO2.
CaCO3(s)
+
CO2(g) +
H2O
====
Ca2+ +
2 HCO3-
a) Derive an expression for the concentration of [H+] and for the [Ca2+]. Using Excel, calculate [H+]
over a range of CO2 pressures (1.0 x 10-5 – 0.01 atm) and use this to plot the [Ca2+] as a function of
PCO2.
b) In what compartment of the aquatic environment might you find CO2 partial pressures as high as
0.01 atm?
Solution:
a) The solubility of calcium carbonate and thus the concentration of Ca2+, will vary with
changes in the PCO2 and [H+]. To derive an expression for the [Ca2+] as a function of PCO2,
we must first determine the [H+].
Rearranging the Ksp expression for CaCO3, we can write;
[Ca2+] = Ksp/[CO32-]
Remembering that we wish to express the [Ca2+] in terms of PCO2 and [H+], we look for an
appropriate substitution for [CO32-].
Rearranging the Ka2 expression gives;
[CO32-] = Ka2 [HCO3-]/[H+]
So we substitute for [HCO3-] using the Ka1 expression;
[HCO3-] = Ka1[CO2(aq)]/[H+]
and we can substitute for [CO2(aq)] using the KH expression;
[CO2(aq)] = KH PCO2
Making these substitutions into the expression for [Ca2+] gives;
K sp [ H +]2
[Ca 2 + ] =
K a1K a 2 K H PCO 2
From this expression we can see that the concentration of calcium depends not only on the
PCO2, but also on the equilibrium concentration of [H+] – which will know will also depend
on the PCO2.
To derive an expression for the [H+] as a function of PCO2, we will need an additional
independent relationship – the charge balance equation. Since all solutions are electrically
solutions assign 3 2006
neutral, we can write that the sum total of positive charges is equal to the sum total of
negative charges.
Thus;
2 [Ca2+] +
[H+]
=
[HCO3-] + 2 [CO32-]
+
[OH-]
For most natural pH ranges, this simplifies to;
2 [Ca2+] = [HCO3-]
since [Ca2+] >> [H+]
and
[HCO3-] >> [CO32-] and [OH-]
So from the charge balance, we can write; [Ca2+] = ½ [HCO3-]
Using the Ka1 and KH expressions to substitute for the [HCO3-] in terms of [H+] and PCO2 (as
before), yields;
[Ca 2 + ] =
K a1K H PCO 2
2[ H + ]
Since we now have two expressions for the concentration of Ca2+ (one derived from the Ksp
expression and the other from the charge balance), we can set them equal to each other and
solve for [H+] as a function of only the PCO2.
Thus;
K sp [ H +]2
K a1K a 2 K H PCO 2
=
K a1K H PCO 2
2[ H + ]
Rearranging and solving for [H+] yields;
2
+
[H ] = 3
2
K a1 K H K a 2 PCO 2
2 K sp
2
Thus, we have an expression to calculate the [H+] at any PCO2. We can use this and then
substitute both the [H+] and the PCO2 into the expression for [Ca2+] above and calculate the
equilibrium concentration of calcium. Using a spreadsheet to repeat this calculation over a
range of PCO2 from 1.0 x 10-5 atm up to 0.01 atm yields the following plot.
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Conc of Calcium in Equilibrium with CaCO3(s) and CO2(g)
PCO2
atm
1.00E-06
2.50E-06
5.00E-06
7.50E-06
1.00E-05
1.25E-05
1.50E-05
1.75E-05
2.00E-05
2.25E-05
2.50E-05
2.75E-05
3.00E-05
4.00E-05
5.00E-05
6.00E-05
7.00E-05
8.00E-05
9.00E-05
1.00E-04
2.00E-04
3.00E-04
4.00E-04
5.00E-04
6.00E-04
7.00E-04
8.00E-04
9.00E-04
1.00E-03
2.00E-03
3.00E-03
4.00E-03
5.00E-03
6.00E-03
7.00E-03
8.00E-03
9.00E-03
1.00E-02
[H+]
M
1.02E-10
1.88E-10
2.98E-10
3.91E-10
4.74E-10
5.49E-10
6.21E-10
6.88E-10
7.52E-10
8.13E-10
8.72E-10
9.29E-10
9.85E-10
1.19E-09
1.38E-09
1.56E-09
1.73E-09
1.89E-09
2.05E-09
2.20E-09
3.49E-09
4.57E-09
5.54E-09
6.43E-09
7.26E-09
8.04E-09
8.79E-09
9.51E-09
1.02E-08
1.62E-08
2.12E-08
2.57E-08
2.98E-08
3.37E-08
3.73E-08
4.08E-08
4.41E-08
4.74E-08
pH
10.0
9.7
9.5
9.4
9.3
9.3
9.2
9.2
9.1
9.1
9.1
9.0
9.0
8.9
8.9
8.8
8.8
8.7
8.7
8.7
8.5
8.3
8.3
8.2
8.1
8.1
8.1
8.0
8.0
7.8
7.7
7.6
7.5
7.5
7.4
7.4
7.4
7.3
[Ca2+]
mM
0.07
0.10
0.13
0.14
0.16
0.17
0.18
0.19
0.20
0.21
0.22
0.22
0.23
0.25
0.27
0.29
0.30
0.32
0.33
0.34
0.43
0.49
0.54
0.58
0.62
0.65
0.68
0.71
0.74
0.93
1.06
1.17
1.26
1.34
1.41
1.47
1.53
1.59
Ka1
Ka2
PCO2
KH(CO2)
Ksp(CaCO3)
4.50E-07
4.70E-11
3.65E-04
3.34E-02
5.00E-09
atm
M atm-1
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Equilibrium [Ca2+] as function of PCO2
1.80
1.60
1.40
[Ca2+] (mM)
1.20
1.00
0.80
0.60
0.40
0.20
0.00
0.0E+00
2.0E-03
4.0E-03
6.0E-03
8.0E-03
1.0E-02
PCO2 (atm)
b) Partial pressures of CO2 of 0.01 atm are found in organic rich sediment pore and
groundwaters. Higher partial pressures of CO2 are found at the sediment interface or in pore
waters at depth below a water column.
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5. Consider dissolve organic matter to have a generic formula {CH2O}. For a water body containing
1.0 mg/L of DOC, calculate the mass in mg of dissolved oxygen in the same volume required to
oxidize it completely. Use this calculation to establish a relation between COD and DOC. Repeat the
calculation using the generic formula for dissolved humic material (Fig 12.3, textbook). Assume
reaction of only the carbon and hydrogen in the humic material.
Solution:
Dissolved Organic Carbon (DOC) is reported in mg/L of carbon whereas Chemical Oxygen
Demand (COD) is reported in mg/L of O2 consumed.
Therefore, if we take {CH2O} as the generic formula of DOC, we can write the
decomposition reaction as follows;
{CH2O} +
O2
Æ
CO2
+ H2O
such that each mole of {CH2O} reacts with one mole of O2. Converting 1.0 mg/L DOC into
moles gives;
1.0 mg/L DOC x 1 mol/12,000 mg = 8.33 x 10-5 mol/L
So in one liter of water the number of moles of O2 consumed is 8.33 x 10-5.
Converting this into a mass in mg gives; 8.33 x 10-5 mol/L x 32,000 mg/1 mol = 2.7 mg/L O2
Thus, 1.0 mg/L DOC corresponds to 2.7 mg/L O2 consumed.
Repeating this calculation with the more complex generic structure for a humic substance
given in Fig 12.3 (text) yields the following decomposition equation (ignoring the oxidation
of nitrogen as directed);
C57H46O18N2
+ 119/2 O2
Æ
57 CO2
+ 23 H2O
So, 1.0 mg/L of DOC = 8.33 x 10-5 M (as before), which we can convert into moles of
generic humic substance as follows;
8.33 x 10-5 mol C/L x 1 mol generic humic/57 mol C = 1.46 x 10-6 mol humic/L
Using this to determine the number of moles of O2 consumed;
1.46 x 10-6 mol humic/L x 59.5 mol O2/1 mol humic = 8.70 x 10-5 mol/L O2
converting this into a mass of O2 consumed yields;
8.70 x 10-5 mol/L O2 x 32,000 mg/1 mol = 2.8 mg/L O2
Thus, refining our relationship between DOC and COD we now get;
1 mg/L DOC = 2.8 mg/L O2 consumed.
solutions assign 3 2006
6. The equivalent weight of humic material is sometimes defined as the mass of humic material that
supplies one mol of binding sites (carboxylic plus phenolic groups). For a particular sample of fulvic
acid (FA) there are 4.2 and 2.2 mmol of carboxylic and phenolic groups per gram of FA, respectively.
What is the equivalent weight?
Solution:
We need to calculate the equivalent weight of this particular fulvic acid (FA), where one
equivalent is defined as the amount that supplies one mole of binding sites. Given the there
are 4.2 x 10-3 mols CO2- per gram of FA and 2.2 x 10-3 mols PhOH per gram of FA, we can
say that the Σ (binding sites) = 6.4 x 10-3 equivalents per gram FA.
Since we want mass per equivalent we use;
1 gram FA
= 156 g/equiv
6.4 x 10−3 equiv
which is the same as saying 156 g per mole of binding sites.
7. Calculate the percent of fulvic acid bound to calcium in a lake sample containing 1.6 mM
of calcium and 12 μg/L of DOM (as fulvic acid). Use the conditional stability constants in
Table 13.3 and assume Ca2+ is the only metal present in significant concentration and a pH
of 5.
Solution: This question requires some knowledge of the association constant (Kf’) between
Ca2+ and fulvic acid, given in Table 13.3 in your textbook and some assumption about the
average concentration of carboxylate groups (CCO2- ~ 5 mmol/g) given in the text.
We need to calculate the percent of fulvic acid bound to calcium, which we can define as
follows;
[Ca − FACO 2 − ]
x100
[ FACO 2 − ]T
where [Ca-FACO2-] represents the concentration of calcium complexed to fulvics and
[FACO2-]T represents the total concentration of fulvic acid carboxylates. We will further
define the following terms; [Ca2+]T , [Ca2+]free and [FACO2-]free
%boundFA =
Such that
[Ca2+]T = [Ca2+]free + [Ca-FACO2-]
and
[FACO2-]T = [FACO2-]free + [Ca-FACO2-]
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From the information provided in the question;
[Ca2+]T = 1.6 x 10-3 M
[FACO2-]T = (12 x 10-6 g/L) x (5 x 10-3 mol/g) = 6.0 x 10-8 M
Since [Ca2+]T >> [FACO2-]T
[Ca2+]T ~ [Ca2+]free
(i.e., the vast majority of Ca2+ is un-bound to FA)
The relevant reaction is given by;
Ca2+ + FACO2- ==== Ca-FACO2-
Kf’ = 1.2 x 103
−
1.2 x10 3 =
[Ca − FACO 2 ]
+
−
[Ca 2 ] free [ FACO 2 ] free
If we let x represent the concentration of complexed fulvic acid ([Ca-FACO2-]),
then the [FACO2-]free = [FACO2-]T - x = 6.0 x 10-8 - x
so
x
1.2 x103 =
−3
(1.6 x10 )(6.0 x10−8 − x)
x = (1.6 x 10-3) (6.0 x 10-8 – x) (1.2 x 103) = 1.15 x 10-7 – 1.92 x
and 2.92 x = 1.15 x 10-7
so x = 3.98 x 10-8 M
Therefore the % bound fulvic = (3.98 x 10-8/6.00 x 10-8) x 100% = 66%
In other words, under the stated conditions, roughly 66% of the fulvic acid carboxylate sites
are bound to calcium.
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8. Draw the structure of the pesticides Fenitrothion and Aldicarb and illustrate how they may
associate with generic humic material (cf. Figs 12.4 – 12.7, textbook). Suggest a location for each of
these pesticides on a triangular air-soil-water plot.
Solution: A quick internet search turned up the following information (structure below).
Fenitrothion
Aldicarb
Class of compound
organothiophosphate
carbamate
CAS #
122-14-5
116-06-3
Empirical Formula
C9H12NO5PS
C7H14N2O2S
Water solubility
30 mg/L
6000 mg/L
Vapour Pressure
18 mPa
13 mPa
Kow
2380
5 - 12
O
-O +
N
O
P
S
O-CH3
S
H3C
H3C
O-CH3
N
CH3
H
N
O
CH3
O
CH3
aldicarb
fenitrothion
Types of humic interactions possible:
Hydrogen bonding with –OH
groups on humics
Electrostatic with metal ion
bridging to O- on nitro group
Electrostatic with O- on humic
and N+ on nitro group
fenitrothion
Hydrogen bonding with NH to
O on humics
Hydrogen bonding with –OH
groups on humics
aldicarb
air
Kow ~ 2400
Kow ~ 10
water sol. ~ 30 mg/L
water sol. ~ 6000 mg/L
vapour press. 13 mPa
soil
water
solutions assign 3 2006