Solution manual of environmental chemistry a global perspective 4th s ch10

a Without calculation, but referring to the appropriate dissociation constants, carefully sketch a single variable species distribution diagram versus pH for this acid in water.. Fill in

PROBLEMS/SOLUTIONS

1 Consider the case of sulfurous acid (H2SO3), which is quantitatively formed when sulfur dioxide gas dissolves

in water

(a) Without calculation, but referring to the appropriate dissociation constants, carefully sketch a single

variable species distribution diagram (versus pH) for this acid in water

(b) Calculate accurately the concentration of all the sulfite species at pH = 7.0

Solution

Part a) (H2O)

SO2 (g) ↔ SO2 (aq) + H2O (l) ↔ H2SO3 (aq)

H2SO3 (aq) ↔ HSO3- + H+ Ka1

HSO3- (aq) ↔ SO32- + H+ Ka2

Look up the Ka values for H2SO3 Ka1 = 1.72 x 10-2 Ka2 = 6.43 x 10-8

(Appendix B.4)

Convert to pKa pKa1 = 1.76 pKa2 = 7.19

(p = -log)

Recall that pH = pKa when there is an equal amount of two adjacent species (conjugated acid/base pair)

Henderson-Hasselbach equation ( pH = pKa + log [conjugate base]/[acid] )

So when [H2SO3] = [HSO3-] in the same solution the pH is 1.76 and α is 0.5, and when [HSO3-] = [SO32-] the pH

is 7.19 and α is 0.5 At very low pH, the fully protonated species will dominate and at very high pH the fully deprotonated form i.e., SO32- will dominate The intermediate species, hydrogen sulfite, HSO3-,will be dominant at

pH = ½ (pKa1 + pKa2) Fill in the lines by joining the identified points to produce an approximate species distribution diagram

82

Sketch the distribution diagram where: ‘y’ axis is α & ‘x’ axis is pH

Note that the pKa values of these species are sufficiently 'far apart' that they can be treated in this simple way Part b)

The amount of SO2 (g) in an unpolluted atmosphere is given in Table 5.2 as the range 0.1 to 0.5 ppbv Assume the concentration is 0.3 ppbv for this calculation

Henry’s Law constant for SO2 (g) is 1.2 x 10-5 mol L-1 Pa-1 (Table 11.1)

0.3 ppbv = 0.00000003% = 3.0 x 10-10 atm = 3.04 x 10-5 Pa

[SO2 (aq)] = 1.2 x 10-5 mol L-1 Pa-1 x 3.04 x 10-5 Pa = 3.65 x 10-10 mol L-1

Assume all SO2 aq is converted to H2SO3 (aq)

[1.0 x 10-7][HSO3-]

Ka1 = - = 1.72 x 10-2 [HSO3-] = 6.3 x 10-5 mol L-1

[3.65 x 10-10]

[1.0 x 10-7][SO32-]

Ka2 = - = 6.43 x 10-8 [SO32-] = 4.0 x 10-5 mol L-1

[6.3 x 10-5]

The total amount of sulfite species present at pH 7.0 is



pH

0

1

1 3 5 7 9 11 13

0.5





SO32- HSO3

-H2SO3

7.19 1.76

3.65 x 10-10 mol L-1 + 6.3 x 10-5 mol L-1 + 4.0 x 10-5 mol L-1 = 1.0 x 10-4 mol L-1

[H2SO3] [HSO3-] [SO32-]

6.3 x 10-5 4.0 x 10-5

αHSO3- = - = 0.63 αSO32- = - = 0.40

1.0 x 10-4 1.0 x 10-4

3.64 x 10-10

αH2SO3 = - = 3.6 x 10-6 ~ 0.00

1.0 x 10-4

total α = 0.63 + 0.40 + 0.00 ~ 1.00 (in reasonable agreement with hand sketched species distribution diagram from part a)

2 Set up a spreadsheet calculation to examine the acid-base speciation of arsenate over the pH range 0 to14 What

is the % distribution of species in acid mine drainage whose pH is 4.9?

Solution

Refer to Appendix B.4 for the dissociation constants for arsenic acid

H3AsO4 Ka1 = 5.8 x10-3 pKa1 = 2.24

H2AsO4- Ka1 = 1.1 x10-7 pKa1 = 6.96

HAsO42- Ka1 = 3.2 x10-12 pKa1 = 11.50

Following the set of Equations (10.12-10.15) set up for phosphate, substitute in the arsenate species Refer also to the instructions given near the end of Section 10.1 on Equilibrium calculations done on spreadsheets, to generate the diagram seen above Each of the three pKa values agree well (visually) with the 0.5 alpha points of intersection for each of these points

At pH 4.9, without calculation, but as viewed on the distribution diagram ~100% of the species will exist as H2AsO4 -

84

3 Use standard thermodynamic relations and assumptions to show that

eq

log ) p

Solution

(Refer to Section 10.2)

Consider the half-reaction

Fe3+ (aq) + e-  Fe2+ (aq)

Using the definition of pE (pE = – log {e-}) and taking logs of both sides, we have

pE = – log ae_ = log Keq + log (aFe / aFe2+)

We also have

ΔGº = – 2.303 RT log Keq = – nFEº

Note that n has the usual electrochemical meaning, i.e the number of electrons transferred in the half-reaction

Therefore, at 298 K and using R = 8.315 J mol-1 K-1 and F = 96 485 C mol-1, we have

In this case, n = 1

eq

Fe

Fe e







2

3

1

K a a

e

eq Fe

Fe







2

n E 0.0591

0.0591

eq

log



a a

Fe Fe



2





Under standard conditions aFe = aFe = 1

aFe3+

log - = 0

aFe

and

For non standard conditions

When the standard pEo value and the actual activities (usually approximated by concentrations) of Fe3+ and Fe2+ are

substituted into this equation, the pE of a particular environmental system can be calculated

In the general case, for a reaction

aA + ne-  bB

where A and B are the oxidized and reduced forms of a redox couple respectively, the reaction quotient (Q) is defined

as

(aB)b

Q = -

(aA)a

The reaction quotient takes the form of an equilibrium constant, but uses activities (or, as an approximation, concentrations) that obtain under any conditions, not just those at equilibrium The general form is then

When pE = 0 and under equilibrium conditions (i.e Keq) then

1

pEº = - log Keq

n

or, this can be expressed as: n(pEº) = log Keq

0.0591

a

Fe

Fe



2





pE = pE - 1



log

86

4 Consider the equilibrium of an aqueous solution containing solid manganese dioxide as indicated in the

following equation

How are the solution’s pH and pE affected by the individual addition of each of the following:

? CaCO and (aq), Fe (g), CO ,

Solution

MnO4- (aq) + 4H3O+ (aq) + 3e-  MnO2 (s) + 6H2O (l)

Addition of: MnO2 (s) - no effect on pH or pE

CO2 (g) - decrease in pH and increase in pE

Fe2+ (aq) - decrease in pE and increase in pH

CaCO3 (s) - increase in pH and decrease in pE

5 A potential measurement in a paddy field yields a reading of –278 mV versus the saturated calomel reference

electrode What is the pE value for this soil?

Solution

E = – 278 mV vs SCE E vs NHE = – 0.278 V + 0.242 V = – 0.036 V

(correction factor for SCE vs NHE)

– 0.036 V

pE = - = – 0.61

0.0591 V

The pE value for the paddy field soil is – 0.61

6 Iron toxicity has occasionally been observed in rice plants grown in lowlands (submerged soils) Use the simple

pE / pH diagram for iron in Fig 10.P.1 to explain how this can happen

Fig 10.P.1 The pE / pH diagram for iron

Solution

The pH of a submerged soil is typically acidic and has a low (0 to - 5) pE value A soil containing large amounts of

any of these following solid forms of iron, Fe(OH)3, Fe(CO3) or Fe2O3 is likely to have Fe2+ (aq) present in

relatively high concentration Iron(II) is stable under these conditions of low pE and slight acidity This could lead

to a situation of iron toxicity

7 Iron(III) hydroxide can act as an oxidizing agent as indicated by the half-reaction

(a) Assuming pH = 7 and otherwise standard conditions, can the oxidation of NH4 to NO3 be effected by this reaction in the hydrosphere?

(b) Could the reaction bring about the oxidation of HS– to SO24 at a pH of 9 (again, under otherwise standard conditions)?

Solution

Part a) There are several ways this question can be answered One is to compare the pE values under the specified conditions To do this, we need to obtain pEo values for each reaction The necessary values for the nitrogen and sulfur reactions are given in Appendix B.5 The value for the iron reaction can be calculated from the tabulated data,

as shown in Section 10.2

NO3- + 10H+ + 8e- ↔ NH4+ + 3H2O pEº = 14.9 Eº = 0.882 V

The calculated pE for a pH of 7 (assuming standard conditions for all other species) is

88

1 [NH4]

pE = pEº - - log -

8 [NO3-][1x10-7]10

pE = 14.9 - 8.75 = 6.2

The pE for the iron system, at pH 7, is calculated based on the following reaction:

Fe(OH)3 (s) + 3H3O+ (aq) + e- ↔ Fe2+ (aq) + 6H2O (l) Eº = 1.005 V

pE = pEº - log ([Fe2+]/[H+]3)

pE = 17 - log (1/(1x10-7) 3) = - 4

When all the components of the two reactions are present together, the reaction whose pE value is more positive is

the reaction that will tend to go from left to right (forward) Therefore, under the conditions specified, nitrate would

be reduced and iron(II) would be oxidized

A second way of solving this problem is to combine the two reactions in a way that indicates iron(III) hydroxide oxidizing ammonium ion to nitrate, as shown

8Fe(OH)3 (s) + 24H3O+ (aq) + 8e- ↔ 8Fe2+ (aq) + 48H2O (l)

NH4 (aq) + 13H2O ↔ NO3- (aq) + 10H3O+ (aq) + 8e

-

8Fe(OH)3 (s) + 14H3O+ (aq) + NH4 (aq) ↔ 8Fe2+ (aq) + 35H2O (l) + NO3- (aq)

If this processes will occur at pH 7, the overall pE value calculated from the two half-reactions should be positive

indicating a spontaneous process

Using Equation 10.48

pE = pEo - 1/n log Q

pE = 17.0 - 14.9 - 1/8 log 1/(10-7)14

= -10.2 The negative sign here indicates that the reaction is not spontaneous as written and would tend to proceed in the reverse direction That is, the Fe(OH)3 will not act as the oxidizing agent, but any NO3- would instead tend to generate NH4 and Fe2+ will precipitate to form Fe(OH)3

Part b)

We will solve this problem, using only the first method shown above

SO42- + 9H+ + 8e- ↔ HS

+ 4H2O pEº = 4.20 Eº = 0.248 V

pE = 4.20 – 1/8 log(1/1x10-9)9

pE = 4.204 – 10.125

pE = – 5.9 (see Figure 10.5)

In this case, the pE value of the reaction is more negative than that for the iron reaction (–4) Therefore, iron(III)

hydroxide is a sufficiently good oxidizing agent to oxidize the HS- species to form sulfate

Note that it is possible to calculate the equilibrium constant for the reactions above, and this will also confirm the conclusions reached by the methods shown

8 Consider the following aqueous uranium species

Calculate the UOH3+ concentration under typical pE and pH conditions for acid mine drainage waters assuming

CU = 1 × 10–5 mol L–1 ΔGfº values / kJ mol–1 are:



2 2

UO – 989.5 UOH3+ – 810.0

U4+ – 579.3

Solution

Typical pE and pH conditions for acid mine drainage are pE = 16 and pH = 3 (see Figure 10.6) Assume a

temperature of 298 K

The reactions being considered are:

1) UO22+ (aq) + 4H3O+ (aq) + 2e- ↔ U4+ (aq) + 6H2O (l) (+ 0.334 V)

2) UO22+ (aq) + 3H3O+ (aq) + 2e- ↔ UOH3+ (aq) + 4H2O (l)

And we have been given the equation:

3) Cu = CU4+ + CUO22+ + CUOH3+ = 1 x 10-5 mol L-1

Solve in terms of CUO22+ for CU4+ and CUOH3+

From Reaction 1) shown above,

ΔGº = (6(–237.18) + (–579.3)) – (4(–237.18 ) + –989.5)) = –64.2 kJ mol-1

pEº = – ΔGº / 2.303 nRT = 5.62

90

and using pE = 16 and pH = 3

1 [U4+]

pE = pEº – - log -

2 [UO22+] [H3O+]4

1 [U4+]

16 = 5.62 – - log -

2 [UO22+] [10-3]4

[U4+]

- = 1.74x10-33

[UO22+]

Calculate ΔGº from given values of ΔGfº for Reaction 2) shown earlier,

ΔGº = (–810.0) + 4(–237.18) – (3(–237.18 ) + (–989.5)) = –57.7 kJ mol-1

pEº = – ΔGº / 2.303 nRT = 5.05 and using pE = 16 and pH = 3

1 [UOH3+]

pE = pEº – - log -

2 [UO22+] [H3O+]3

1 [UOH3+]

16 = 5.05 – - log -

2 [UO22+] [10-3]3

[UOH3+]

- = 1.26x10-31

[UO22+]

[UOH3+] = 1.26x10-31 [UO22+] [U4+] = 1.74x10-33 [UO22+]

Substitute known relationships into Equation 3 (in terms of UO22+) and solve

Equation 3) CU4+ + CUO22+ + CUOH3+ = 1 x 10-5 mol L-1

1.74x10-33 [UO22+] + CUO22+ + 1.26x10-31 [UO22+] = 1 x 10-5 mol L-1

(0 + 1 + 0)[UO22+] = 1 x 10-5 mol L-1

[UO22+] = 1.0 x 10-5 mol L-1

The uranium species will end up entirely as the UO22+ under the conditions of acid mine drainage

The UOH3+ concentration will be essentially zero (It is calculated to be 1.26x10-36 mol L-1)

9 Figure 10.P.2 is a partially completed pE / pH diagram for arsenic species

Fig 10.P.2 The pE / pH diagram for some arsenic species

(a) Calculate the equation for the line (shown) for the boundary H2AsO4 (aq) / H3AsO3 (aq)

(b) Calculate equations for the lines (not shown) for the boundaries H3AsO4 / H2AsO4 and H3AsO4 /

H3AsO3

Assume a temperature of 25ºC For boundaries use Cspecies = 1 × 10 –4 mol L–1

(c) Comment on predicted As species in seawater and acid mine drainage In both situations assume that the solution is well aerated ΔGfº values / kJ mol–1 are:

H3AsO4 – 769.3



4

2AsO

H – 748.8

H3AsO3 – 640.0



3

2AsO

H – 587.7

H2O – 237.2

Solution

Part a) The reaction being considered is:

H2AsO4 - (aq) + 2e- + 3H+ (aq)  H3AsO3 (aq) + H2O (l)

ΔGº = (– 640.0 + –237.2) – (–748.8 + 0 + 0) = – 128.4 kJ

92

128 400 J

pEº = - = 11.3

11 413.07

1 [H3AsO3]

pE = pEº – log -

2 [H2AsO4-][H+]3

Using the boundary condition of 1 x 10-4 mol L-1

3 1

pE = 11.3 – - log - pE = 11.3 – 1.5pH

2 [H+]

Part b) Consider H3AsO4 / H2AsO4- boundary – no redox

H3AsO4 (aq)  H2AsO4 - (aq) + H+ (aq)

ΔGº = (–748.8 + 0) – (–769.3) = 20.5 kJ

– 20 500 J

log K = - = – 3.6

5706.53

[H2AsO4-][H+]

K = - at the boundary [H3AsO4] = [H2AsO4-] = 10-4 M

[H3AsO4]

K = [H+] log K = log [H+] = – 3.6

pH = 3.6 is the boundary condition (depends only on pH)

Part b) Consider H3AsO4 / H3AsO3 boundary – redox condition

2 electron process

H3AsO4 (aq) + 2 e- + 2 H+  H3AsO3 (aq) + H2O

ΔGº = – 107.7 kJ

107 700 J

pEº = - = 9.44

11 413.07

1 1

pE = 9.44 – - log -

2 [H+]2

pE = 9.44 – pH

Part c) The type of arsenic species that would be found in acid mine drainage would be the higher oxidation state

(As-V) and it is likely to be fully protonated (H3AsO4) The seawater species will also be the higher oxidation state, but likely deprotonated (H2AsO4 -)

10 As we have mentioned in the text, Eh is sometimes used in place of pE

(a) Calculate the lower and upper stability boundary lines for water using Eh

(b) Show that if you divide each line determined in part (a) by 0.0591 that you end up with the

equations shown in the text (pE = ̶ pH and pE = 20.80 – pH, for the lower and upper

boundaries respectively)

(c) Using Fig 10.P.1, the iron pE/pH diagram in problem 6, convert the scale of pE shown to

Eh What must you do?

Solution

Part a) 2H2O (l) + 2e-  H2 (g) + 2 OH- (aq) E° = – 0.828V (the lower boundary)

Use the Nernst equation:

Eh = E° – (0.0591/n) log(PH2 (g) x [OH-]2 / 1)

n = 2, PH2 (g) = 1.00 atm, (note: pOH + pH = 14, or pOH = 14 – pH)

Eh = – 0.828V – (0.0591/2) log(1.00 x [OH-]2 / 1)

The two ‘2’s cancel (one inside log term (exponent) and the two outside log term (denominator))

Eh = – 0.828V – (0.0591/1)log(1.00 x [OH-] / 1)

Eh = – 0.828V – (0.0591)log[OH-]

Eh = – 0.828V + (0.0591) pOH (note the change in sign)

Eh = – 0.828V + (0.0591)(14 – pH)

Eh = – 0.828V + 0.828V – 0.0591 pH

Eh = – 0.0591 pH