ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
4-12
2.5*10-7 mole of NH3(aq) and 2.5*10-7 mole NaOH are added to 1 liter of distilled water at
25°C. What is the pH if μ=0?
Charge Balance
Na + + NH 4+ + H + = OH −
[
] [
] [ ] [
]
This condition is satisfied where
pH=7.77
3
4
5
6
pC
7
8
`
9
10
11
12
13
0
2
4
6
8
pH
1 of 19
10
12
14
16
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
4-13
10-2 moles NH4Ac and 10-2 moles NaOH are added to 1 liter of water at 25°C. What is
the pH if μ=0?
Charge Balance
Na + + NH 4+ + H + = OH − + Ac −
[
] [
] [ ] [
] [ ]
pH=10.6
0
1
2
3
pC
4
`
`
5
6
7
8
9
0
2
4
6
8
pH
2 of 19
10
12
14
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
4-14 pH of a Solution Having Two Bases by Graphical Method
What is the pH of a 25°C solution containing 10-3 moles NaHCO3 and 2x10-3 mole NH3
per liter if μ=10-2?
Write the proton condition, where proton reference levels are HCO3- and NH3
[H 2CO3 ] + NH 4+ + H + = OH − + CO32 −
[
] [ ] [
] [
]
The equilibrium constants need to be adjusted for ionic strength. New equilibrium
constants based on concentration, cK, may be calculated.
For μ=10-2, from Fig 3-4 find the activity coefficients using the Güntelberg
Approximation.
γ1±=0.9; γ2±=0.66
{ }{
[ ]
}
[
K w = H + OH − = γ H + H + γ OH − OH −
For the carbonate system
{H }{HCO } = γ [H ]γ
=
+
K a1
H+
{H 2 CO3 }
HCO3−
c
[HCO ]
γ H CO [H 2 CO3 ]
2
2−
3
−
3
+
H+
2−
3
CO32−
For the ammonia system
+
H + {NH 3 } γ H + H γ NH 3 [NH 3 ]
=
K a1 =
NH 4+
γ NH + NH 4+
[ ]
4
[
c
]
Kw =
Kw
c
Ka2 =
K a1 =
10−14
= 10−13.91
0.9 × 0.9
γ H CO K a1 1 × 10−6.3
= 10− 6.2
K a1 =
=
γ H γ OH
0.9 × 0.9
2
γ HCO K a 2
−
3
γ H γ CO
+
c
=
γ H +γ OH −
3
+
−
3
HCO3−
{ }
{ }
−
3
3
{H }{CO } = γ [H ]γ [CO ]
=
[HCO ]
{HCO }
γ
+
Ka2
+
−
3
]
2−
3
γ NH K a1
+
4
γ H γ NH
+
3
=
−
0.9 × 10−10.3
= 10−10.1
0.9 × 0.66
0.9 × 10−9.3
=
= 10− 9.3
0.9 × 1
Construct the appropriate pC-pH diagram using these modified coefficients, note
CT,CO3=10-3M and CT,NH3=2*10-3M
To find the solution pH, make the simplifying assumptions, [NH4+]>>[H2CO3]>>[H+] which
reduces the proton condition to
[NH ] = [OH ] + [CO ]
+
4
−
2−
3
Find the ammonium concentration to intersect the sum of the hydroxyl ion and carbonate
ion concentration at
pH = 9.85
3 of 19
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
0
1
2
3
pC
4
`
`
5
6
7
8
9
0
2
4
6
8
pH
4 of 19
10
12
14
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
4-15
A nitric acid solution, pH=2.7, results from NOx removal from a stack gas. Neglect ionic
strength effects and the temperature is 25°C.
a. How much Na2CO3 must be added to neutralize this solution prior to discharge?
(The final pH is 8.3. Assume that no weak acids are present in the scrubber
water.) Hint: What is the predominant carbonate species at pH 8.3?)
b. What is the buffer intensity of the final solution?
At pH 2.7, almost all of CT,NO3 is present as NO3-. Therefore, the neutralization reaction is
CO32-+H+ÆHCO3-.
a. Calculate the quantity of H+ to be neutralized,
H+=10-2.7-10-8.3=10-2.7
Add 10-2.7 moles Na2CO3 per liter of discharge
b. Calculate the buffer intensity, β, of the final solution using Eq 4-99
β = 2.3([H + ] + [OH − ] + α 0α1CT ,CO 3 + α1α 2CT ,CO 3 )
[H+]=10-8.3 [OH-]=10-5.7
CT,CO3=10-2.7
For the carbonate system at pH=8.3
α0=0.01
α1=0.98
α2= 0.01
Neglect H+ as insignificant
β = 2.3(10−5.7 + 0.01 × 0.98 × 10−2.7 + 0.98 × 0.01 × 10−2.7 ) = 9.5 × 10−5
β=9.5*10-5M
This is the buffering capacity of the solution
5 of 19
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
4-16 Neutralization of a Strong Acid with a Strong Base
How much soda ash, Na2CO3, in moles/liter is required to neutralize a pickle liquor
solution containing 10-1 mole H2SO4/liter? Assume that the H2SO4 will react only with
the Na2CO3 and that the pH after neutralization is 8.3.
The neutralization reaction is
H++CO32-ÆHCO3H+=2*10-1M and will be practically 100% neutralized in being reduced to 10-8.3M,
therefore,
2*10-1M Na2CO3
Supply this much soda ash
6 of 19
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
4-17 pH Buffer for Breakpoint Chlorination
An experiment is to be conducted on breakpoint chlorination (the oxidation of ammonia
by chlorine) in which it is desired to maintain the pH constant within 0.5 units of the
initial pH of 8. Make the assumption that all of the ammonia is in the NH4+ form (a
reasonable assumption since the pKa of NH4+ is 9.3 and the pH of interest is pH 8). The
breakpoint reaction between Cl2 and NH3 proceeds as follows:
3Cl 2 + 2 NH 4+ ↔ N 2( g ) + 8 H + + 6Cl −
The maximum amount of ammonia that will be used in the experiments is 12.5 mg as
N/liter (0.89*10-3 moles/liter). Select an appropriate acid-conjugate base pair and
determine the concentration of it that will control the pH to within 0.5 units of 8.0 during
the reaction. Neglect ionic strength effects; the temperature=25°C. Determine the buffer
intensity of this solution. (Note: There is a detailed discussion of the breakpoint
chlorination reaction in Chapter 7.)
a. Select the dihydrogen phosphate (H2PO4-), monohydrogen phosphate ion (HPO42-)
pair, as it has the closest pKa to 8.0 of species in Table 4-1, excluding hypochlorite.
Now, to determine the quantity of the acid-base pair to add to provide the desired control.
The worst case condition is when 8.9*10-4M of NH4+ are added, producing 3.56*10-3M
H+.
So, find CT, that will absorb that many equivalents of H+, given the knowledge of the
species of phosphate at pH 7.5 and 8.0.
α 2, pH =7.5 × CT − α 2, pH =8 × CT = 3.56 × 10 −3
(0.86 − 0.67 )× CT
= 3.56 ×10 −3
Solve for CT
CT=1.87*10-2M
Add this much phosphate salt (as HPO42-) to buffer pH, adjust
to pH=8.0 to obtain proper starting distribution of species.
b.Calculate β
Use Eq 4-99 β = 2.3( H + + OH − + α 0α1CT ,CO 3 + α1α 2CT ,CO 3 )
[ ] [
]
α2=0.863
At pH=8.0
α0=0 α1=0.137
Neglect all terms as insignificant, except the last one
β = 2.3(0.137 × 0.863 × 1.87 × 10 −2 ) =5.09*10-3 M
β=5.09*10-3 M
α3=0
This is the buffering capacity of the solution
7 of 19
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
4-18
An industrial wastewater is to be discharged to a receiving stream with a pH of 8.3 and a
total alkalinity =2*10-3 eq/liter. The wastewater contains 5*10-3 M H2SO4, and the pH of
the stream should not be permitted to drop below 6.3.
a. What is the maximum dilution ratio (volume waste/volume stream water) that can
be used for discharge of the wastewater?
b. What is the buffer intensity of the solution at pH 6.3?
a. At pH=8.3, nearly all the carbonates are in the form of HCO3-. So
[HCO3-]=2*10-3M. At pH 6.3, (pKa1), the carbonates would be half H2CO3 and
half HCO3-). So, 1*10-3M of bicarbonate would be neutralized. Thus, 1*10-3M
of H+ can be added per liter of stream water. The wastewater is a strong
diprotic acidconcentrated at 5*10-3M. Thus, it has 1*10-4M H+.Thus, no more
than 100 ml of wastewater should be added per liter of stream water.
The dilution ratio is 0.1
b. The buffer intensity is calculated according to Eq 4-99
β = 2.3( H + + OH − + α 0α1CT ,CO 3 + α1α 2CT ,CO 3 )
[ ] [
]
At pH=6.3, neglect H+ and OH-, α2 is also negligible.
β = 2.3(α 0α 1CT ,CO 3 )
β = 2.3(0.5 × 0.5 × 2 ×10 −3 )
β=1.15*10-3M
This is the buffering capacity of the solution
8 of 19
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
4-19
A sample of natural water contains 1*10-3 M CO32- and 3*10-3 M HCO3-.
a. As the pH is lowered during the alkalinity titration, at which pH is the CO2 in
solution in equilibrium with atmospheric CO2?
b. What is the pH of the total alkalinity equivalence point, pHCO2? (Neglect dilution
effects.) Give your answer to the nearest 0.1 pH unit.
c. If 50 percent of the CO2 formed during the titration “escapes,” what is the “new”
pH of the total alkalinity equivalence point?
a. This would be the pH where [H2CO3]=10-5M.
1
[H 2CO3 ] = α o × CT ,CO3 =
× 4 × 10 −3 = 10 −5
− 6.3
− 6.3
−10.3
10
10 × 10
1+
+
+
2
H
H+
[ ]
+
-9
[ ]
[H ]=1.12*10 M
pH=8.9
b. pHCO2=0.5*(pKa1+pCT,CO3)
pCT,CO3=-log(1*10-3M+3*10-3)=2.40
pHCO2=0.5*(6.3+2.4)=4.35
pHCO2=4.35
c. Recalculate pCT,CO3=-log(0.5*4*10-3)=2.70
pHCO2=0.5*(6.3+2.7)=4.50
pHCO2=4.50
9 of 19
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
4-20
A sample of natural water that has been equilibrated with CaCO3(s) is isolated from its
surroundings. Indicate whether the addition of small quantities of the following will
increase, decrease, or have no effect on the total alkalinity or total acidity and state very
briefly why. Neglect ionic strength effects.
a.
b.
c.
d.
e.
HCl
FeCl3
Na2SO4
CO2
Na2CO3
Species
HCl
FeCl3
Na2SO4
CO2
Total Alkalinity
Decrease
Decrease
No effect
No effect
Na2CO3 Increase
Total Acidity
Increase
Increase
No effect
Increase
No effect
Comment
H+ is a strong acid
Ferric iron is a strong acid
Sulfate a very weak base
CO2 is the TAlk endpoint, therefore
cannot include in the expression
CO32- is the TAcid endpoint, therefore
cannot include in the expression
For example, when carbonic acid (CO2(aq)) is added to water it has no net affect on total
alkalinity, because it produces a hydrogen ion and bicarbonate ion. Each cancels the
effect of the other.
CO2 + H 2 O ↔ H 2 CO3 + H 2 O ↔ H + + HCO3−
It can similarly be shown, that soda ash when added to water has no net effect on total
acidity.
10 of 19
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
4-21 Alkalinity Titration, and pH of Buffered Water Receiving Strong
Base Waste
A natural water has the following partial analysis:
pH=8.3
[Ca2+]=5*10-4M
-3
[HCO3 ]=3*10 M
[Mg2+]=1*10-4M
-5
[CO2(aq)]=3*10 M
[SO42-]=1*10-4M
a. What volume of 0.02N H2SO4 is required to titrate a 100-ml sample to the total
alkalinity endpoint? What is the total alkalinity in eq/liter and in mg/liter as
CaCO3?
b. A waste containing 10-2 moles NaOH/liter is to be discharged to this water. The
pH cannot be raised above 9.5. What is the maximum number of liters of waste
that can be added to each liter of the natural water?
a. Total alkalinity consists of titrating all bases. In the given water, HCO3-, is the only
significant base present for total alkalinity. Sulfate is far too weak a base and would not
neutralize before the total alkalinity endpoint.
Alk = 3*10-3eq/liter = 150 mg/l as CaCO3
The volume of 0.02eq/liter H2SO4 required per liter to provide 3*10-3 eq is
3 *10 −3
eq
eq 1
= V × 0.02
l
l 100 ml
Add V=15 ml of 0.02 N H2SO4 to titrate to the total alkalinity endpoint
b. At pH=8.3, nearly all the carbonates are in the form of HCO3-. So
CT,CO3=3*10-3M. At pH 9.5, [HCO3-]=α1*CT,CO3=0.863*3*10-3=2.59*10-3M
The amount of bicarbonate neutralized by base is then,
3*10-3-2.59*10-3=4.1*10-4M
This is the number of equivalents of base that may be added per liter of natural
water. Use this information to find the volume of waste.
1 l waste
eq
l waste
×
= 0.041
l natural 0.01 eq
l natural
V= 41 ml of waste per liter of natural water
4.1× 10 − 4
11 of 19
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
4-22 Acidity and Alkalinity
A partial water analysis is given as follows:
CO2 = 44 mg/liter
[Cl-]=1*10-3M
[HCO3-]=2*10-3M
[SO42-]=1*10-4M
a. What is the solution pH and CO32- concentration?
b. What is the caustic, carbonate, and total alkalinity (in eq/liter and in mg/liter as
CaCO3)?
c. What is the mineral, CO2, and total acidity (in eq/liter)?
d. What is the pH if 5*10-4 mole OH- (as NaOH) are added per liter of the above
sample?
mg mmol
×
= 1× 10 −3 M
l
44mg
= [CO2 ] + HCO3− = 1× 10 −3 + 2 × 10 −3 = 3 × 10 −3
a. CO2 = 44
CT ,CO 3
[
]
[CO2 ] = 1×10 −3 M =
1
1
× CT ,CO 3 =
× 3 ×10 −3 = 1× 10 =3
−6.3
K
10
1 + a+
1+
H
H+
[ ]
α0 =
1
3
[ ]
pH=6.6
K a1 K a 2
[CO ] = α
2−
3
2 × CT ,CO 3 =
[H ]
+ 2
K
K K
1 + a+1 + a1 a22
H
H+
× CT ,CO 3 = 1.33 × 10 −4 × 3 ×10 −3
[ ] [ ]
[CO32-]= 3.99*10-7M
pH=6.6
b. From Tb 4-11
Caustic Alkalinity
Kw
10 −14
+
− H − CT ,CO 3 (α 1 + 2α 0 ) = −6.6 − 10 −6.6 − 3 × 10 −3 × (0.33 + 2 × 0.67 )
+
H
10
Caustic Alkalinity = negative therefore = 0, reasonable since pH < 10.3
Carbonate Alkalinity
K
10 −14
CT ,CO 3 (α 2 − α 0 ) + w+ − H + = 3 × 10 −3 × (1.33 × 10 −4 − 0.33) + −6.6 − 10 −6.6
H
10
Carbonate Alkalinity = negative therefore = 0, reasonable since pH < 8.3
[ ] [ ]
[ ] [ ]
Total Alkalinity
CT ,CO 3 (α 1 + 2α 2 ) +
[ ] [ ]
Kw
10 −14
+
−3
−4
−
H
=
×
×
+
×
×
+
− 10 −6.6
3
10
0
.
67
2
1
.
33
10
− 6.6
+
H
10
(
)
Neglect pH terms
Total Alkalinity = 0.002 eq/liter = 100 mg/l as CaCO3
12 of 19
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
c. From Tb 4-11
Mineral Acidity
K
10 −14
H + − w+ − CT ,CO 3 (α 1 + 2α 2 ) = 10 −6.6 − −6.6 − 3 ×10 −3 (0.67 + 2 ×1.33 × 10 − 4 )
H
10
Mineral acidity = negative therefore = 0, reasonable since pH > 4.5
[ ] [ ]
CO2 Acidity
[ ] [HK ] = 3 ×10 × (0.33 − 1.33 ×10 )+ 10
CT ,CO 3 (α 0 − α 2 ) + H + −
−3
w
+
−4
−6.6
−
10 −14
10 −6.6
CO2 acidity=9.9*10-4 eq/liter
Total Acidity
[ ] [HK ] = 3 ×10
CT ,CO 3 (α 1 + 2α 0 ) + H + −
w
+
−3
× (0.67 + 2 × 0.33) + 10 −6.6 −
10 −14
10 −6.6
Total acidity=4*10-3 eq/liter
d. Adding 5*10-4 moles of OH-, the hydroxyl ions react preferentially with CO2
Thus, after neutralization of the added OH[CO2]=1*10-3-5*10-4=5*10-4
[HCO3-]=2*10-3+5*10-4=2.5*10-3
Calculate pH from α0
5 × 10 −4
1
α0 =
=
= 0.167
10 −6.3 3 × 10 −3
1+
H+
[H+]=10-7
pH=7.0
[ ]
13 of 19
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
4-23 Titration for Species Determination
Fifty ml of a natural water sample is titrated with 0.02 N H2SO4. The titrant volume
required to titrate to pH 8.3 is 6 ml and an additional 8 ml is required to titrate to pH 4.3.
a. What is the caustic, carbonate, and total alkalinity in meq/liter?
b. What is the mineral, CO2, and total acidity in meq/liter?
c. What is the CT,CO3?
d. What is the pHCO2 and pHCO32- (to the nearest 0.1 pH unit)?
e. What is the [H+], [OH-],[CO32-],[HCO3-] and [H2CO3*] in the original sample?
Calculate using both the approximate method described in Section 4.13.3 and the
exact procedure and compare the results.
a. The caustic alkalinity is 0 meq/liter, since Vmo>Vp implying that HCO3- was present
in solution.
Carbonate alkalinity = Vp*N/V= 6 ml * 0.02 eq/l / 50 ml
Carbonate alkalinity = 2.4 meq/l
Total alkalinity = Vmo*N/V= 14 ml * 0.02 eq/l / 50 ml
Total alkalinity = 5.6 meq/l
b. Mineral acidity and CO2 acidity are zero since initial pH>8.3
Total acidity = bicarbonate originally present = (Vmo-Vp)*N/V = (8-6)ml*0.02 eq/l/ 50 ml
Total acidity=0.84 meq/l
c. The total carbonate concentration is the sum of the bicarbonate and carbonate ions
originally present in the solution. Or by Eq 4-118, it is the difference between the total
alkalinity and the carbonate alkalinity. The total alkalinity counts CO32- twice, subtracting
the carbonate alkalinity corrects for this.
5.6-2.4=3.2
CT,CO3=3.2 meq/l
d. pHCO2=0.5*(pKa1+pCT,CO3)=0.5*(6.3+2.49)
pHCO2=4.40
pHCO32-=14-0.5*(pKb1+pCT,CO3)=14-0.5*(3.7+2.49)
pHCO32-=10.91
e. Approximate Method (Tb 4-9)
pH=10.3+log(2.4/0.84)=10.76
pOH=3.24
[CO32-]=2.4*10-3M (from a)
[HCO3-]=8.4*10-4M (from b)
[H2CO3*] =0
Exact Method
-
2-
-3
CT,CO3=[HCO3 ]+[CO3 ]=3.2*10 M
Charge Balance
[H ][CO ]
=
[HCO ]
+
K a2
[H+]=[OH-]+[HCO3-]+2[CO32-]
14 of 19
2−
3
−
3
[ ][
K w = H + OH −
]
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
Make substitutions
10 −14
H+ =
+ α 1CT ,CO 3 + 2α 2 CT ,CO 3
H+
⎛
10 −6.310 −10.3
10 −6.3
⎜
2
10 −14 ⎜
H+
H+
+
H =
+
+2
10 −6.3 10 −6.310 −10.3
H + ⎜⎜ 10 −6.3 10 −6.310 −10.3
1
1+
+
+
+
2
2
⎜
H+
H+
H+
H+
⎝
David E. MacNevin
[ ] [ ]
[ ]
[ ] [ ]
+
[ ]
-11
[ ]
[ ]
[ ]
[ ]
[H ]=2.82*10
pH=10.55
pOH=3.45
2[CO3 ]=α2*CT,CO3=0.64*3.2*10-3=2.05*10-3 M
[HCO3-]= α1*CT,CO3=0.36*3.2*10-3=1.15*10-3 M
[H2CO3*] = α0*CT,CO3=2.02*10-5*3.2*10-3=6.46*10-8 M
15 of 19
⎞
⎟
⎟
−3
⎟3.2 × 10
⎟
⎟
⎠
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
4-24 Effect of Ionic Strength on Carbonate Equilibrium
A solution has a carbonate alkalinity of 1 meq/liter and a total alkalinity of 6 meq/liter.
The ionic strength of the solution is 10-2 M and the temperature is 40°C. Calculate [H+],
[HCO3-], and [CO32-] neglecting ionic strength effects and compare the values obtained
when corrections are made for ionic strength effects. Use the constants given in Tables 42 and 4-7 for your calculations.
When μ=0 and T=40°C
Kw=2.95*10-14=10-13.53
Ka1=10-6.30
Ka2=10-10.22
For μ=10-2, from Fig 3-4 find the activity coefficients using the Güntelberg
Approximation.
γ1±=0.9; γ2±=0.66
{ }{
}
[ ]
Kw
=
[
K w = H + OH − = γ H + H + γ OH − OH −
c
Kw =
γ H + γ OH −
2.95 × 10 −14
= 10 −13.44
0.9 × 0.9
For the carbonate system
{H }{HCO } = γ [H ]γ
=
+
K a1
H+
{H 2 CO3 }
HCO3−
[HCO ]
γ H CO [H 2 CO3 ]
2
2−
3
−
3
+
H+
2−
3
CO32−
HCO3−
−
3
c
γ H CO K a1 1× 10 −6.3
=
= 10 −6.21
0 .9 × 0 .9
γ H γ OH
K a1 =
2
−
3
3
+
3
{H }{CO } = γ [H ]γ [CO ]
=
[HCO ]
{HCO }
γ
+
Ka2
+
−
3
]
c
K a2 =
−
γ HCO K a 2
−
3
γ H γ CO
+
2−
3
=
0.9 ×10 −10.22
= 10 −10.04
0.9 × 0.66
Calculate the analytical carbon concentration (use Eq 4-118)
CT,CO3=Total Alkalinity-Carbonate Alkalinity=6*10-3-1*10-3=5*10-3eq/liter
From Tb 4-11
Total Alkalinity=6*10-3M
CT ,CO 3 (α 1 + 2α 2 ) +
Kw
H+
[ ]
In general
K a1 K a 2
⎛
K a1
⎜
2
+
H
H+
⎜
− [H ] = 5 × 10 −3 ⎜
+2
K
K K
K
K K
1 + a+1 + a1 a22
⎜⎜ 1 + a+1 + a1 + a22
H
H
H
H+
⎝
[ ]
[ ]
[ ] [ ]
[ ] [ ]
When μ=0
16 of 19
⎞
⎟
⎟ Kw
+
⎟+ H+ − H
⎟⎟
⎠
[ ] [ ]
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
⎛
10 −6.3010 −10.22
10 −6.30
⎜
2
⎜
H+
H+
2
+
TotAlk = 5 × 10 −3 ⎜
− 6.30
10 −6.30 10 −6.3010 −10.22
10 −6.3010 −10.22
⎜ 1 + 10
1
+
+
+
2
2
⎜
H+
H+
H+
H+
⎝
[ ]
[ ]
+
-10
[H ]=2.82*10
[ ]
[ ]
[ ]
pH=9.55
[ ]
David E. MacNevin
⎞
⎟
⎟ 10 −13.53
− H+
⎟+
+
H
⎟
⎟
⎠
[ ] [ ]
When μ=0
−6.30
10
10 −9.55
[HCO3-]=α1*CT,CO3=
× 5 × 10 −3
− 6.30
− 6.30
−10.22
10
10 10
1 + −9.55 +
2
10
10 −9.55
[HCO3-]=4.1*10-3M
When μ=0
[
[
]
]
[
]
10 −6.3010 −10.22
[10 ]
−9.55 2
2-
[CO3 ]=α2*CT,CO3=
1+
− 6.30
− 6.30
−10.22
10
10 10
+
−9.55
10
10 −9.55
[
]
[
[CO32-]=9.2*10-4M
× 5 × 10 −3
]
2
When μ=0
When μ=10-2
⎛
10 −6.2110 −10.04
10 −6.21
⎜
2
⎜
H+
H+
TotAlk = 5 × 10 −3 ⎜
+
2
− 6.21
10 −6.21 10 −6.2110 −10.04
10 −6.2110 −10.04
⎜ 1 + 10
+
+
+
1
2
+
+
⎜
+ 2
H
H
H
H+
⎝
[ ]
[ ]
+
-10
[H ]=3.98 *10
[ ]
[ ]
[ ]
pH=9.40
-2
[ ]
⎞
⎟
⎟ 10 −13.44
− H+
⎟+
+
H
⎟
⎟
⎠
[ ] [ ]
When μ=10
−6.21
10
10 −9.40
× 5 ×10 −3
[HCO3-]=α1*CT,CO3=
−6.21
−6.21
−10.04
10
10 10
1 + −9.40 +
2
10
10 −9.40
[HCO3-]=4.1*10-3M
When μ=10-2
[
[
]
]
[
]
10 −6.2110 −10.04
[10 ]
−9.40 2
[CO32-]=α2*CT,CO3=
1+
[CO32-]=9.3*10-4M
− 6.21
− 6.21
−10.04
10
10 10
+
−9.40
10
10 −9.40
[
]
[
× 5 × 10 −3
]
2
When μ=10-2
Increased ionic strength caused pH to decrease slightly, while HCO3- remained
unchanged, and CO32- showed a small increase in concentration.
17 of 19
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
David E. MacNevin
4-25 pH of a Poorly-Buffered Lake Receiving Acid Rain
In some regions carbonate-bearing minerals are lacking in the earth and the lakes in these
regions have a low alkalinity. Acid rains, resulting from conversion of industrial SO2
emissions to H2SO4 , can cause significant pH depressions in such lakes and may result in
fish kills. What volume of acid rain, pH =4.0, is required to lower the pH of a lake to 6.7?
The lake has a volume of 20*106 ft3, a pH =7.0, and an alkalinity of 30 mg/liter as
CaCO3.
Some alkalinity is present in the water. Assume it is all present as bicarbonate alkalinity,
since pH<8.3 Assume T=25°C.
Calculate CT,CO3 at pH=7.0
K a1
10 −6.3
10 −7.0
H+
=
= 0.833
α1 =
K a1 K a1 K a 2
10 −6.3 10 −6.310 −10.3
1+ + +
1 + −7.0 +
2
2
H
10
H+
10 −7.0
[ ]
CT ,CO 3
[
[ ] [ ]
[
[HCO ] = 6 ×10
=
α1
−
3
]
] [
]
−4
0.833
-4
CT,CO3=7.20*10 M
Calculate [HCO3-] at pH=6.7
K a1
10 −6.3
10 −6.7
H+
=
= 0.715
α1 =
K a1 K a1 K a 2
10 −6.3 10 −6.310 −10.3
1+ + +
1 + −6.7 +
2
2
H
10
H+
10 −6.7
[ ]
[ ] [ ]
[
[
]
] [
]
[HCO3-]=α1CT,CO3=0.715*7.20*10-4=5.148*10-4M
Thus, in every liter of water the amount of bicarbonate alkalinity destroyed (and
corresponding H+ demand) will be
6*10-4-5.14*10-4=8.52*10-5M
Note: The reaction of interest for the buffer is, H++HCO3-ÆH2CO3
Also for every liter of water, H+ will be needed to lower the pH.
10-6.7-10-7=9.95*10-8M
Together, for each liter of water, the H+ required to change from pH 7.0 to pH 6.7 is
8.52*10-5+9.95*10-8
8.53*10-5M (or N)
Now calculate the number of equivalents required over the entire lake
18 of 19
ENV 6519
Chapter 4 HW Part 2 (#12-#25)
June 14, 2006
The lake has volume 2*107ft3, convert to liters.
7.48 gal 3.78l
2 × 10 − 7 ft 3 ×
×
= 5.655 × 108 l
3
gal
ft
Over the entire lake that number is
(5.655*108)*(8.53*10-5)=4.82*104 equivalents
The rain has a concentration 1*10-4 N
⎡ ft 3 ⎤
4.82 ×10 4 eq
8
4
.
82
10
l
=
×
×
⎢
⎥
1×10 −4 N
⎣ 28.27l ⎦
V=1.71*107ft3
19 of 19
David E. MacNevin