CE4501 Environmental Engineering Chemical Processes
Problem Set 3 - SOLUTION
Fall 2008
Due: Monday, 10/13 by 5 p.m.
DO 10 OF THE 12 PROBLEMS.
Solutions will be posted on the Web. Problem sets will be graded for completeness, and one
problem (selected at random) will be graded in detail. Each problem set contributes 2.5%
towards your final grade.
1. Calculate the pH of the following solutions. Ignore activity corrections. SOLUTION: a. 10‐5 M Ca(OH)2 (strong base) This is a strong based that dissociates completely to yield 2x10‐5 mole/L of OH‐ in solution; therefore, pOH is 4.7, and pH = 14 – pOH = 9.3. (Note that activity correction would change this by less than 0.03 pH units.) b. 10‐3 M HClO4 (strong acid) This is a strong acid that dissociates completely to yield 0.001 mole H+ /L; therefore, pH is –log(0.001) = 3. (Note that activity correction would change this by less than 0.03 pH units.) c. 10‐3 M HOCl (pKa = 7.5) This is a weak acid, and therefore we cannot assume that 0.001 Mole/L of H+ are released into solution. The actual concentration of H+ will be less than this, so the pH will be > 3.0. There are 3 unknowns (H+, OCl‐, HOCl), so we need three equations (mass balance, Ka, and charge balance). To calculate the actual conc., rearrange the mass balance equation to: [HOCl] = CT – [OCl‐] Substitute this into the pKa equation and rearrange to solve for [OCl‐]: H + }{OCl − } { H + } γ 1 ⎡⎣OCl − ⎤⎦
{
=
Ka =
CT − [OCl − ]
{HOCl}
CT K a
−
[OCl ] =
{H + } γ 1 − K a
(
)
Now, we substitute into the charge balance equation so there is only one unknown, {H+}: [H+] = [OCl‐] + [OH‐] Because we added an acid and pH will be below 7, we can neglect [OH‐] [ H + ] = [OCl − ] =
CT K a
({ H } γ
+
1
)
+ Ka
γ 12 ⎡⎣ H + ⎤⎦ + K a ⎡⎣ H + ⎤⎦ − CT K a = 0
+
[H ] =
+
[H ] =
−Ka +
( Ka )
− 4γ 12 ( −CT K a )
2
2γ 12
−10−7.5 +
(10 ) + 4 ( 0.96 ) 10
2
−15
2 ( 0.96 )
−3
⋅10−7.5
2
[ H + ] = 10−5.23
pH = − log ({H + }) = − log ( γ 1[ H + ]) = 5.25
d. 10‐3 M NaOCl (pKa = 7.5) We can begin with exactly the same 2 steps as in part c above. The charge balance equation contains Na+ whose concentration we know to be 10‐3 M. [H+] +[Na+] = [OCl‐] + [OH‐] We can substitute as we did above: [ H + ] + [ Na + ] = [OCl − ] + [OH − ]
{H + }
γ1
+ CT =
CT K a
({ H } γ
+
1
{H + }2 + CT ⋅{H + }γ 1 =
+ Ka
)
+
Kw
{H + }γ 1
CT K a ⋅{H + }γ 1
({ H } γ
+
1
+ Ka
)
+ Kw
{H + }3 γ 1 + {H + }2 K a + {H + }2 γ 12CT + {H + } ⋅ CT K aγ 1 = CT K a ⋅{H + }γ 1 + {H + }γ 1 K w + K w K a {H + }3 γ 1 + {H + }2 ( K a + γ 12CT ) + {H + } ( CT K aγ 1 − CT K aγ 1 − γ 1 K w ) − K w K a = 0
{H + } = 10−9.23
pH = 9.23
‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 2. For a 1x10-3 M Na2S solution (pKa1 = 7.1, pKa2 = 18.0):
a. Find the pH at equilibrium;
SOLUTION: It is easiest to use a pC-pH diagram for this problem.
pH
0
2
4
6
8
10
12
14
0
2
4
pOH
pC
6
pH
pH2S
8
pHSpS2-
10
12
14
The proton reference level is S2-, H2O, and the proton condition is:
2[H2S] + [HS-] + [H+] = [OH-]
It is easiest to find this point by following the [OH-] line to where it intersections with the [HS-]
line at pH 11 (dashed line).
b. At equilibrium, which is larger, [HS-] or [S2-]?
SOLUTION: The diagram above shows clearly that [HS-] is about 7 orders of magnitude larger
than [S2-].
3. Problem 11.5 in the text (p. 225). Line 3 should read: (TOC) < 250 mg/L as C.
An industry is discharging an acetic acid waste stream. Its NPDES permit requires pH > 4 and
TOC < 250 mg/L as C. What is the maximum strength of acetic acid that can be discharged?
SOLUTION:
We must compare the two constraints imposed by the permit and find the lower of the two.
TOC < 250 mg/L as C:
mgC 1 mmole CH 3COOH
250
⋅
= 10.4mM acetic acid
24mgC
L
pH > 4:
To find the acetic acid concentration that will produce pH of 4, recall that one H+ is generated
for every HAc that dissociates to produce Ac-.
{H + }{ Ac − }
{H + }2
Ka =
=
{HAc}
CT − {H + }
{H + }2 + K a {H + } − K a CT = 0
CT =
{H + }2 + K a {H + }
Ka
CT = 0.0006 M = 0.6mM
The pH constraint requires a lower acetic acid concentration than does the TOC limit, and hence
the maximum allowable acetic acid concentration is < 0.6 mM or 14.4 mg C/L
4. Problem 11.15 in the text (p. 228). Hint: use a pC-pH diagram.
Find the total carbonate concentration of a pure Na2CO3 solution so that [HCO3-] = [CO32-].
SOLUTION:
This can be solved either algebraically or with a pC-pH diagram. I’ll show both solutions.
Because we know the solution will be basic, we can assume that {H+} and [H2CO3] are
negligible. Therefore:
CT = [H2CO3] + [HCO3-] + [CO32-] ~ [HCO3-] + [CO32-] ~ 2[HCO3-] = 0.5[Na+]
The charge balance equation is:
[H+] + [Na+] = [HCO3-] + [OH-] + 2[CO32-] or
2CT = [OH-] + 0.5CT + 2·0.5CT
Î 0.5CT = [OH-]
Now, we can figure out pH and pOH using the Ka equation:
{H + }{CO32− } {H + } ⋅ 0.5CT
Ka =
=
{HCO3− }
0.5CT
{H + } = K a = 10−10.3
Therefore, pOH = 14 - 10.3 = 3.7
Using the relationship 0.5CT = [OH-] we can calculate CT to be 2[OH-] or 2x10-3.7M = 0.4 mM
Using the pC-pH diagram shown below, we see that if we want [HCO3-] = [CO32-] we must be at
pH = pKa. This immediately fixes the pH to 10.3. The proton condition for a Na2CO3 solution
is:
[H+] + 2[H2CO3] + [HCO3-] = [OH-]
At pH 10.3, both of the first terms are neglible so the proton condition becomes:
[HCO3-] = [OH-]
pH
0
2
4
6
8
10
12
14
0
2
4
pOH
6
pC
pH
H2CO3
HCO3
8
CO3
10
12
14
Thus, all that must be done is to ove the carbonate lines down (lower CT) until the OH‐ line intersects the HCO3‐ line at the pKa or the point where HCO3‐ and CO32‐ lines cross. This is shown to happen below at a pCT of 3.4 pH
0
2
4
6
8
10
12
14
0
2
4
pOH
pC
6
pH
H2CO3
8
HCO3
CO3
10
12
14
5. Ammonium is a weak acid/base system with a pKa of 9.3 (ΔHo= +52.21 kJ/mole). NH4+ <==> NH3 + H+ Ammonium is a byproduct of all sewage treatment plants; organic nitrogen is converted to ammonium, and only a fraction of this is oxidized to nitrate in conventional treatment plants. Discharge of ammonium to receiving waters is problematic both because of the potential oxygen consumption through oxidation of the NH4+ to nitrate and because of ammonia (NH3) toxicity to fish. These problems are particularly acute in Onondaga Lake in New York State. a. The toxicity of ammonia is influenced strongly by pH. The carbonate system is the primary control on pH in most lakes. Would you expect the pH of the hypolimnion to be higher or lower than the pH of the epilimnion of a productive lake? Why? ANSWER: The pH of the hypolimnion is probably lower than that of the epilimnion because aerobic respiration in the hypolimnion release considerable CO2 which hydrolyzes to form carbonic acid that in turn lowers the pH. This is partially offset by release of bicarbonate in zones where anaerobic respiration occurs. b. The total concentration of NH4+/NH3 (i.e., CT‐NH3 = [NH4+] + [NH3]) typically is higher in the hypolimnion than in the epilimnion; NH4+ is produced in the sediments from the decomposition of settled algal biomass. Would you expect the greatest NH3 toxicity problems in the hypolimnion or in the epilimnion? Why? ANSWER: It is difficult to predict where the problem will be greatest. In the hypolimnion, the total ammonium concentration is higher, but the pH is lower resulting in a larger fraction being in the nontoxic form, NH4+. The higher pH in the epilimnion results in a larger fraction being in the toxic form, but because the total concentration is lower, the net difference may be small. c. In August of 1988 the following measurements were made in Onondaga Lake: Epilimnion Hypolimnion CT‐NH3 2 mg N/L 10 mg N/L pH 8.2 7.5 o
8oC Temp. 25 C Did the concentration of NH3 exceed the Final Chronic Value (FCV) for fish of 0.03 mg/ N/L in the hypolimnion? ANSWER: We must first calculate the Ka for 8oC using the value of ΔH given above.
⎛ K ⎞ ΔH ⎛ 1 1 ⎞
ln ⎜ 2 ⎟ =
⎜ − ⎟
R ⎝ T1 T2 ⎠
⎝ K1 ⎠
kJ
1 ⎞
⎛ 1
mol
ln ( KT =8o ) − ln (10−9.3 ) =
−
⎜
⎟
kJ
⎝ 298.15 281.15 ⎠
8.314 x10−3
mol ⋅ K
ln ( KT =8o ) = −21.414 − 1.274 = −22.688
52.21
KT =8o = 10−9.85
Now we may use this equilibrium concentration and the specified pH and CT to find [NH3]:
{NH 3 }{H + }
{NH 3 }{H + }
Ka =
=
{NH 4+ }
( CT − [ NH 3 ]) γ 1
[ NH 3 ] =
K a CT
{H } + K a γ 1
+
1mmolN 1mmolNH
⋅
⋅
(10 ) ⎛⎜10 mgN
L 14mgN 1mmolN
−9.85
[ NH 3 ] =
⎝
3
⋅10−3
mol ⎞
mmol ⎟⎠
10−7.5 + 10−9.85 ⋅ 0.93
mol
mgN
mgN
[ NH 3 ] = 3.18 x10−6
⋅14, 000
= 0.044
L
molNH 3
L
The concentration in the hypolimnion (0.044 mgN/L) does exceed the FCV for fish (0.03
mgN/L).
------------------------------------------------------------------------------------------------------
6. Problem 12.2 in the text (p. 265).
Sketch the titration curves for the titration of 0.01 M Na2CO3 and 0.001 M Na2CO3 with strong
acid.
SOLUTION:
To sketch the curves qualitatively, one needs only 3 points for each: the initial pH (g=0), the
pKa2 (pH at g = 0.5), the first (bicarbonate) equivalence point (g = 1.0), pKa1 (pH at g = 1.5) and
the endpoint pH (g = 2.0). To calculate the curves quantitatively, one uses the charge balance:
[Na+] + [H+] = [OH-] + [HCO3-] + 2[CO32-] + CA
Rearranging and substituting the alpha values for bicarbonate and carbonate:
CA = [Na+] + [H+] – [OH-] – α1CT - 2α2CT
Recall that
K a1[ H + ]
α1 = + 2
[ H ] + K a1[ H + ] + K a1 K a 2
K a1 K a 2
[ H ] + K a1[ H + ] + K a1 K a 2
Substituting these into the charge balance one obtains an equation for CA that depends only on
pH (H+ concentration):
K
C K [H + ]
2CT K a1 K a 2
− + 2
C A = 2CT + [ H + ] − w+ − + 2 T a1 +
[ H ] [ H ] + K a1[ H ] + K a1 K a 2 [ H ] + K a1[ H + ] + K a1 K a 2
The plot of Ca vs. pH (the titration curve) for each value of CT is shown below.
α2 =
+ 2
12
0.01M
0.001M
10
pH
8
6
4
2
0
0
0.005
0.01
0.015
0.02
Ca (eq/L)
A How do the pH values at the zeroth, first, and second equivalence points change with the total
carbonate concentration?
ANSWER: The pH values at two of the three equivalence points change with increasing CT.
E.g., the zeroth equivalence points fall at pH = 10.6 and 11.1 for 0.001 and 0.01 M carbonate,
respectively. The pH values at the second equivalence point change from 4.05 (0.01M) to 4.6
(0.001M). The pH values at the first equivalence point remain the same at 8.3.
B. Where are the regions of high buffer intensity? What species are responsible for the high
buffer intensity in each region?
ANSWER: The regions of high buffer intensity are in the vicinity of pH = pKa1 and pH = pKa2.
At the lower pKa, it is bicarbonate that is important; at the higher pKa, carbonate ion is the major
buffer.
7. Problem 12.7 in the text (p. 265).
An industry discharges a waste stream consisting of a monoprotic acid HX. The titration curve
for the titration of the waste stream with NaOH is shown in the text. Unfortunately, the
technician forgot to record the HX concentration in the waste stream, the NaOH concentration of
the titrant, or amount of base added.
A. What is the approximate pKa of HX?
ANSWER: The pKa is about 7.1, the region of greatest buffering.
B. What is the total HX concentration in the waste stream?
ANSWER: Because we cannot read the x axis scale, we have to use the pH at the endpoint (~10)
and the proton condition to deduce to total concentration. The proton condition for this endpoint
pH is:
[H+] + [HX] = [OH-]
Because [H+] is negligible, [HX] = [OH-] = 10-4 M
From the Ka equation we can determine [X-] and add it to [HX] to obtain CT.
Ka = [H+][X-]/[HX] Î [X-] = Ka[HX]/[H+] = (10-7.1)(10-4)/(10-10) = 10-1.1
CT = [HX] + [X-] = 10-4 + 10-1.1 = 10-1.099 M
8. Problem 13.3 in the text (p. 293).
The pH of a lake water is measured in the field to be 8.0. A 100 mL sample of the water
required 5.0 mL of 0.10 N HCl to titrate it to the H2CO3 equivalence point.
A. What is the alkalinity of the water?
ANSWER: The alkalinity is the meq of acid required divided by the sample (+ acid) volume:
Alkalinity = (5.0 mL)(0.1eq/L)/(100 + 5.0 mL) = 0.00476 eq/L
B. What are the concentrations of CT, H2CO3, HCO3- and CO32- in the lake?
ANSWER: We may take the alkalinity to be equal to the HCO3- concentration which at pH 8 is
very nearly equal to CT. The two Ka equations may be used to calculate H2CO3 and CO32-.
Ka1 = [H+][HCO3-]/[H2CO3] Î [H2CO3] = [H+][HCO3-]/Ka1
[H2CO3] = (10-8)(0.00476)/10-6.3 = 9.5x10-5 M
Similarly,
Ka2 = [H+][CO3-]/[HCO32-] Î [CO32-] = [HCO32-]Ka2/[H+]
[CO32-] = (0.00476)(10-10.3)/(10-8) = 2.4x10-5 M
9. Problem 13.8 in the text (p. 293).
PollutiCon Inc., a large multinational widget manufacturer, wishes to discharge 50,000 gallons of
0.1 M H2SO4 into a lake. The lake has an alkalinity of 100 mg/L as CaCO3, a pH of 7.5, a mean
depth of 7 ft, and a surface area of 3 acres. What will be the pH of the lake water after the
addition of the waste?
SOLUTION:
Before calculating the pH, it is useful to compare the alkalinity in units of meq/L with the
concentration of acid once it is added to the lake.
⎞ ⎛ meq ⎞
mg
mmole
meq
⎛
⎞⎛
Alk = ⎜100
as CaCO3 ⎟ ⎜
⎟⎜ 2
⎟=2
L
L
⎝
⎠ ⎝ 100mgCaCO3 ⎠ ⎝ mmole ⎠
⎛
L ⎞ ⎛ mole ⎞ ⎛ eq ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ acre ⎞ ⎛ ft ⎞ ⎛ −3 m3 ⎞
Acid = ( 50, 000 gal ) ⎜ 3.78
⎟
⎟ ⎜ 0.1
⎟⎜
⎟⎜ 2
⎟⎜
⎟⎜
⎟⎜
⎟ ⎜10
gal ⎠ ⎝
L ⎠ ⎝ mole ⎠ ⎝ 7 ft ⎠ ⎝ 3acre ⎠ ⎝ 4047 m 2 ⎠ ⎝ 0.305m ⎠ ⎝
L ⎠
⎝
eq
Acid = 0.0015
L
The comparison shows us that the acid is equal to 75% of the total alkalinity. While the pH will
not be titrated down to the pH 4.3 endpoint, it will be titrated past the g = 0.5 mark at which
point the pH will equal the pKa of 6.3. This comparison alone is enough to show that the
companies claims are false. To obtain an exact value for the pH, we would use the charge
balance equation to determine pH when Ca equals 0.0015 eq/L
Kw
CT K a1[ H + ]
2CT K a1 K a 2
+
− + 2
C A = CT + [ H ] − + − + 2
+
[ H ] [ H ] + K a1[ H ] + K a1 K a 2 [ H ] + K a1[ H + ] + K a1 K a 2
A plot of Ca vs. pH is shown below. We can read from the graph that at Ca = 0.0015, pH = 5.8.
9
8
7
6
pH
5
4
3
2
1
0
0
0.0005
0.001
0.0015
0.002
0.0025
0.003
0.0035
Ca (eq/L)
10. Problem 13.10 in the text (p. 294).
a. If a small amount of NaCl is added to a natural water, what happens to the alkalinity?
ANSWER: It stays the same. Na+ and Cl- do not appear in the terms that contribute to
alkalinity (alk = [HCO3-] + 2[CO32-] + [OH-] – [H+]).
b. If 5 mL of 0.1 M NaOH and 5 mL of 0.1 M H2SO4 are added, what happens to alkalinity?
ANSWER: Because the acid is twice as strong as the base, there is a net addition of acidity (H+)
and the alkalinity will decrease. In the alkalinity definition, [H+] is subtracted; therefore, an
increase in [H+] will cause a decrease in alkalinity.
c. If the sample is titrated to the H2CO3 endpoint, what happens to the alkalinity?
ANSWER: The alkalinity is reduced to zero.
d. Titrate it to the second equivalence point (f = 2)?
ANSWER: To reach the second equivalence point (f=2) base must be added. Addition of base
will increase the alkalinity.
11. Problem 13.12 in the text (p. 294).
Create a plot showing the lime dose required (mg/L) to raise the wastewater pH to 11 as a
function of the initial alkalinity of the water (mg/L as CaCO3) over the normal alkalinity range of
wastewaters (50-200 mg/L as CaCO3). Assume an initial pH of 7.5.
SOLUTION:
Again, the easiest approach is to use the charge balance equation to calculate the amount of base
required:
[H+] + [Cationswaste] + 2[Ca2+] = [OH-] + [HCO3-] + 2[CO32-]
[H+] + Alk + 2CT-lime = Kw/[H+] + α1Alk + 2α2Alk
CT-lime = 0.5(Kw/[H+] + α1Alk + 2α2Alk – [H+] – Alk)
⎛ K
⎞
C
K [H + ]
2C
K K
CT −lim e = 0.5 ⎜ w+ + + 2 T − alk a1+
+ + 2 T − alk a+1 a 2
− [ H + ] − CT − alk ⎟
⎝ [ H ] [ H ] + K a1[ H ] + K a1 K a 2 [ H ] + K a1[ H ] + K a1 K a 2
⎠
The plot of lime dose vs. alkalinity is shown below.
180
160
Lime dose (mg/L)
140
120
100
80
60
40
20
0
0
50
100
150
200
250
Initial Alk (mg/L as CaCO3)
12. An intriguing mystery is developing with respect to the ion composition of the effluent
from the wastewater treatment plant (WWTP). The bar charts below contrast the ion
concentrations in the WWTP sample and the Houghton tap water sample. There are three
major differences between the two samples: 1) Concentrations of Na+ and Cl- are about
four times higher in the WWTP sample; 2) the sulfate concentration in the WWTP is 28
times higher than in the Tap water; and 3) the bicarbonate ion concentration (calculated
here as the difference between the sum of cations and anions) is about 5 times lower in
the WWTP sample. In this problem, let us focus on the sulfate and bicarbonate ions.
Could the measured ion concentrations and pH values have resulted from the addition of
strong acid (H2SO4) to the WWTP sample? What would be the bicarbonate ion
concentration and pH of the tap water if it were titrated with 2,150 μeq/L of H2SO4?
Assume that the system is open and in equilibrium with air with a CO2 concentration of
370 μatm. (Suggestion: There are many ways in which you could approach this
problem. One method is to construct a pC-pH diagram showing the tap water alkalinity
(2.85 meq/L) as Ca2+, the WWTP sulfate concentration (2.15 meq/L), and an opensystem for the carbonate ions in equilibrium with the air. The proton condition is
difficult to derive, so it is easier to work with the charge balance equation. What are the
pH and HCO3- concentration at the point where the charge balance equation is satisfied
on the pC-pH diagram?)
Cations
WWTP
Na+, ClK+, NO3Mg2+, SO42Ca2+, HCO3-
Anions
0
1000
2000
3000
4000
5000
6000
7000
8000
C on c . ( m e q / m 3 )
Cations
Tap w ater
Na+, ClK+, NO3Mg2+, SO42Ca2+, HCO3-
Anions
0
1000
2000
3000
4000
5000
6000
7000
8000
C o nc . ( m e q / m 3 )
SOLUTION:
The charge balance equation is:
[H+] + [Cations] = [Other anions] + [SO42-] + [HCO3-] + 2[CO32-] + [OH-]
[SO42-] = [Catons] – [Other anions] – [Alk]
The pC-pH diagram for the tap water above indicates that if the system were open to the
atmosphere, the tap water should have had a pH of 8.7; since we measured a pH of 7.86, either
we did not measure all ions correctly or the sample must have been somewhat supersaturated
with CO2 (the excess H2CO3 lowered the pH below 8.7).
The pC-pH diagram for the waste water sample does indicate that it should have a lower
equilibrium pH of about 7.8. The pH that we measured was 6.6. There is not enough sulfuric
acid in the sample to have lowered the pH down to pH 6.6 if the sample is in equilibrium with
the atmosphere.
We could also calculate a titration curve for the sample just as we did in the problems above.
The titration equation is:
K
C K [H + ]
2CT K a1 K a 2
− + 2
C A = CCation − Anion + [ H + ] − w+ − + 2 T a1 +
[ H ] [ H ] + K a1[ H ] + K a1 K a 2 [ H ] + K a1[ H + ] + K a1 K a 2
The Ccation-anion represents the concentration of cations (Na+, K+, Ca2+, Mg2+) minus the sum of
non-acid anions (Cl-, NO3-) which in the tap water amounts to 2.86 meq/L. We can use this
equation to calculate the amount of acid that we would need to add to the tap water to obtain the
measured pH of 6.65.
This titration curve suggests that it would take 0.9 meq/L of acid to titrate the tap water sample
to pH 6.65. There is a total of 2.15 meq/L more SO42- in the waste water as compared with the
tap water. Therefore, it is possible that addition of sulfuric acid could have lowered the pH to the
measured value of 6.6. Evidently, not all of the sulfate was added as sulfuric acid or the pH
would have been lowered to approximately 5.6. The two approaches do not yield contradictory
information; they simply indicate that the system was not in equilibrium with the atmosphere,
and for that reason the first method is not accurate.