Download CE4501 Environmental Engineering Chemical Processes

CE4501 Environmental Engineering Chemical Processes
Problem Set 4 - SOLUTION
Fall 2008
1. Problem 15.3 in the text (p. 334).
Using the formation constants in section 15.5.2, find the concentrations of the hydroxo complexes of Cu
(including the dimer) in a water at pH 6.0 with total soluble copper equal to 1.0x10-6M
SOLUTION:
We are asked to find the concentrations of CuOH+, Cu(OH)2o, Cu(OH)3-, Cu(OH)42- and Cu2(OH)22+.
The text gives the stability constants as β values on p. 323:
Reaction
log β
Symbol
2+
+
Cu + OH ÍÎ CuOH
6.3
β1
Cu2+ + 2OH- ÍÎ Cu(OH)2o
12.8
β2
2+
14.5
β3
Cu + 3OH ÍÎ Cu(OH)3
Cu2+ + 4OH- ÍÎ Cu(OH)4216.4
β4
2+
2+
2Cu + 2OH ÍÎ Cu2(OH)2
17.7
We know that the ligand (OH-) concentration is fixed at 10-8 M, so we can use the approach that [CuLn]
= βn[Cu2+][OH-]n because the ligand concentration will not change as a result of the complexation. The
mass balance equation for Cu then becomes:
CT-Cu = [Cu2+] + [CuOH+] + [Cu(OH)2o] + [Cu(OH)3-] + [Cu(OH)42-] + 2[Cu2(OH)22+]
CT-Cu = [Cu2+](1 + β1[OH-] + β2[OH-]2 + β3[OH-]3 + β4[OH-]4 + [Cu2+]K22[OH-]2)
Species
Cu2+
CuOH+
Cu(OH)2o
Cu(OH)3Cu(OH)42Cu2(OH)22+
CT-L
log(β)
10-8
10-8
10-8
10-8
10-8
6.3
12.8
14.5
16.4
17.7
βn[L]n
(1)
10-1.7
10-3.2
10-9.5
10-15.6
10-12.3
% of total Cu
98%
1.96%
0.06%
<0.01%
<0.01%
<0.01%
Conc. (mole/L)
0.98 μM
0.02 μM
6.2x10-4 μM
3.1x10-12 μM
2.5x10-22 μM
4.9x10-19 μM
2. Problem 15.7 in the text (p. 335).
Calculate the change in the HCN concentration in a cyanide bath at pH 7 and CNT = 1x10-4 M if 1x10-5
M total Ni(II) is added.
SOLUTION:
The basic idea is that the nickel with complex with much of the cyanide and leave less cyanide available
to become protonated (and thus volatile and toxic). The possible forms of cyanide and CN-, HCN, and
Ni(CN)42-. The possible forms of Ni in addition to the cyanide complex are Ni2+, NiOH+, Ni(OH)2o,
Ni(OH)3-T. First, let’s calculate the concentration of HCN in the absence of Ni(II):
CT = [HCN] + [CN-] Î [CN-] = CT – [HCN]
K a = 10
−9.2
+
{H + }{CN − } {H } ( CT − [ HCN ])
=
=
{HCN }
[ HCN ]
{H + }CT
10−7 ⋅10−4
=
= 9.9 x10−5
K a + {H + } 10−9.2 + 10−7
As expected, at pH well below the pKa (9.2), most of the cyanide is in the form of HCN.
[ HCN ] =
To determine the amount of HCN when the Ni is present, we could either set up 7 equations to solve for
7 unknowns or we can try to simplify the situation. Because each Ni(II) ion can bind with 4 CN- ions,
the metal could consume up to 40% of the total cyanide; also because the pH is < pKa, we cannot
assume that the ligand concentration (CN-) is much greater than the metal concentration. However, we
can calculate the pKa1 for Ni and find it to be (4.1-14 =) 9.9; this means that at pH 7, the hydroxide
complexes are not important. This allows us to reduce the system to 4 unknowns (Ni2+, Ni(CN)42-, CN-,
HCN) and 4 equations:
CT-Ni = 10-5 = [Ni2+] + [Ni(CN)42-]
CT-CN = 10-4 = [HCN] + [CN-] + [Ni(CN)42-]
Ka = [H+][CN-]/[HCN]
K4 = [Ni(CN)42-]/[Ni2+][CN-]4
Rearranging:
[Ni2+] = CT-Ni – [Ni(CN)42-]
[HCN] = [H+][CN-]/Ka
[Ni(CN)42-] = K4[CN-]4CT-Ni/(1 + K4[CN-]4)
Substituting
CT-CN = 10-4 = [H+][CN-]/Ka + [CN-] + K4[CN-]4CT-Ni/(1 + K4[CN-]4)
⎛
K 4 [CN − ]4 CT − Ni ⎞
C
−
⎜ T −CN
⎟
1 + K 4 [CN − ]4 ⎠
⎝
−
Rearranging: [CN ] =
[H + ]
+1
Ka
Solving by iteration we find that [CN-] = 5.64x10-7 or 10-6.25
From the Ka relation, we can then find [HCN] to be 8.94x10-5 M = 10-4.05
Therefore, the change in HCN concentration is 9.9x10-5 – 8.94x10-5 or 9.56x10-6 M. In other words, the
HCN concentration was reduced about 10%.
Verification of assumptions:
1. Hydroxo complexes of Ni are unimportant:
At pH 7 and free Ni2+ of 6.22x10-11, β1[OH-][Ni2+] = 104.110-710-10.2 = 10-13.1M, a value much
lower than the NiCN4 complex.
-----------------------------------------------------------------------------------------------------------------3. Problem 15.13 in the text (p. 336).
A community uses a groundwater supply for drinking water. The water is fairly hard with a hardness of
300 mg/L as CaCO3. The minimize soap scum, the residents wish to add a detergent containing NTA to
their washing machines. The detergent is 3% NTA by weight. If the average washing machine has a
capacity of 40 L, how much detergent should be added to reduce the hardness to 24 mg/L as CaCO3?
Assume all hardness is from calcium. The molecular weight of NTA is 188 g/mole.
SOLUTION:
The total Ca2+ concentration is 300 mg/L as CaCO3 or 3 mM. Of this, 2.75 mM must be bound by the
ligand such that only 0.25 mM of hardness remains:
CT-Ca = 3x10-3 M = [Ca2+] + [CaL-]
The mass balance above assumes that the carbonato complex and the ternary complex with NTA (CaL24) are unimportant. We will have to verify those assumptions later.
NTA is a triprotic acid with pKa’s of 2.0, 2.94 and 10.334; thus, we should probably consider the HL2and L3- forms in the mass balance for ligand:
CT-L = [HL2-] + [L3-] + [CaL-]
The unknowns are: Ct-L, [HL2-], [L3-] and [H+]. We have four unknowns, but because it is unlikely that
the pH is determined by the detergent (we do not even know the form in which the NTA is added), I
shall assume a pH of 7.5. For the 3 unknowns, we have the mass balance equation above and two
equilibrium equations:
Ka3 = {H+}{L3-}/{HL2-}
β1 = {CaL-}/{Ca2+}{L3-} = 106.3
Rearranging: [Ca2+] = CT-Ca – [CaL-] = 2.5x10-4
Substituting into the stability constant equation: [L3-] = [CaL-]/[Ca2+]β1 = (2.75x10-3)/(2.5x10-4)106.3
Î [L3-] = 5.5x10-6 M
Now we can solve for HL
[HL2-] = {H+}[L3-]/Ka = 10-7.5/10-10.334·[L3-] = 682·[L3-] = 3.76x10-3
The total NTA required is therefore: CT-L = [HL2-] + [L3-] + [CaL-] = 3.76x10-3 + 5.5x10-6 + 2.75x10-3
Or CT-L = 6.52x10-3
To get the amount of detergent, we need to account for the volume of the washing machine (40 L) and
the concentration of concentrated detergent (3%):
Vwm ⋅ CT − L = Vadded ⋅ Cdet ergent
(40 L) ( 6.52 x10−3 M )
Vadded
= 1.6 x103 g
mole
3 gNTA
⋅
100 g det ergent 188 gNTA
This is an implausible amount of detergent to add.(~ 1 L) and might indicate that a higher pH would be
more reasonable.
V ⋅C
= wm T − L =
Cdet ergent
Verify assumption:
1. CaL2 is unimportant
2. H2L and H3L are unimportant
4. Draw a pC‐pH distribution diagram for the Cu species listed below. Cu2+ + OH‐ ÍÎ CuOH+ log β = 6.3 o
2+
‐
log β = 11.8 Cu + 2OH ÍÎ Cu(OH)2 Cu2+ + 4OH‐ ÍÎ Cu(OH)42‐ log β = 16.4 2+
2+
‐
log β = 17.7 2Cu + 2OH ÍÎ Cu2(OH)2 Cu2+ + 2OH‐ ÍÎ Cu(OH)2(s) log β = 19.3 SOLUTION: The first step is to put the equilibrium constants into the form needed for a pC‐pH diagram. *K values can be interpreted as equivalent to Ka values. First we calculate Kn as βn/βn‐1 and next *K is calculated as K∙Kw. The dimer does not fit easily into the calculation scheme and will be omitted. Reaction log β log(K) log(*K) pKa Cu2+ + OH‐ ÍÎ CuOH+ 6.3 6.3 ‐7.7 7.7 7.7 o
2+
‐
Cu + 2OH ÍÎ Cu(OH)2 11.8 5.5 ‐8.5 8.5 14.5 2.7 ‐11.3 11.3 18.8 Cu2+ + 3OH‐ ÍÎ Cu(OH)3‐ 2‐
2+
‐
16.4 1.9 ‐12.1 12.1 12.1 Cu + 4OH ÍÎ Cu(OH)4 2+
2+
‐
2Cu + 2OH ÍÎ Cu2(OH)2 17.7 NA NA 19.3 13.0 ‐1 1 Cu2+ + 2OH‐ ÍÎ Cu(OH)2(s) The first graph shows the result using the soluble Cu(OH)2o and the second graph shows the result using the much less soluble Cu(OH)2(s). Note that the dimer doesnot really lend itself to constructing an acid dissociation constant and is therefore not included in the analysis. The differences between these two models will be discussed in more detail when we reach the precipitation/dissolution topic. 5.a. Which metal, Cu2+ or Mn2+, would you predict to bind more strongly (i.e., have a higher binding constant) with Cl‐? Why? ANSWER: Cl‐ is a hard ligand that would bind preferentially with Mn2+ because it is a harder metal than Cu2+. Therefore the binding constant for MnCl+ should be larger than that for CuCl+. b. Which metal, Ca2+ or Hg2+, would you predict to bind more strongly (i.e., have a higher binding constant) with the N and S groups in cysteine? Why? 2+
(cysteine) ANSWER: As in part A, a soft metal (Hg ) will bind more strongly with a soft ligand than will a hard metal (Ca2+) if the ion charges are the same. 6. In utilizing stability constants, one must be certain to which reaction they apply or exactly how they are defined. To calculate the extent of formation of CaHCO3+ it would be easiest to use the equilibrium constant for the reaction: Ca2+ + HCO3‐ <==> CaHCO3+ However, what one would find in reference texts are the constants for the following reaction: Ca2+ + H+ + CO32‐ <==> CaHCO3+ log K = 11.59 How would you calculate the equilibrium constant for the first reaction above if you had only the value for the second reaction and the pKa (10.3) for the dissociation of HCO3‐? What is the value of the equilibrium constant for the first reaction above? SOLUTION: We must arrange the equations needed to obtain the desired net reaction. The log K values may then be added. Ca2+ + H+ + CO32‐ <==> CaHCO3+ log K = 11.59 log K = ‐10.3 HCO3‐ ÍÎ CO32‐ + H+ ‐
+
2+
log K = 1.29 Ca + HCO3 ÍÎ CaHCO3 7. The major inorganic ligands in Torch Lake are HCO3‐ (1 mM), SO42‐ (50 μM), and OH‐ (10‐5.9M). The total concentration of Ca2+ in the lake is 500 μM, and the pH is 8.1. Calculate the fraction of the total Ca2+ that is bound as CaHCO+ and CaSO4o. The formation constants are given below. Ca2+ + HCO3‐ <==> CaHCO3+ log K = 1.3 2+
‐
+
Ca + OH <==> CaOH log K = 1.15 Ca2+ + SO42‐ <==> CaSO4o log K = 2.31 SOLUTION: Because we know that only a small fraction of the ligands will bind, we can use the approach of calculating βa[L]a. Complex log(β) CT‐L βa[L]a Percentage +
‐3
‐2
CaHCO3 1.3 10 2x10 1.94% CaOH+ 1.15 10‐5.9 1.8x10‐5 <0.01% CaSO4o Ca2+ 2.31 5x10‐5 1.02x10‐2 The percentages are calculated as % =
0.99% 97.07% β a [ L]a
1 + ∑ β a [ L]a
8. EDTA is a quatroprotic, hexadentate ligand which is used in some countries as a binding agent in detergents. Unlike NTA, EDTA is not biodegraded in sewage treatment plants nor in the environment. It is slowly photodegraded in lakes, but high concentrations would be expected to build up if it were widely used. To examine some of the environmental implications of the use of EDTA, we will examine its acidity characteristics, trace metal and major cation binding abilities. A) Plot a pC‐pH diagram describing the distribution of EDTA species over the pH range of 0‐14. Assume a total concentration of 10‐5 M. The four pKa values are: 2.72, 3.24, 6.68, and 11.12. SOLUTION: This is a straightforward acid pC‐pH diagram. The diagram shows that HL3‐ is the dominant form of EDTA within the range of surface water pH values. B) Given a total Ca2+ concentration of 500 μM, and a pH of 8.0, calculate the fraction of Ca2+ bound by EDTA. Ignore all other complexes of Ca2+ other than CaEDTA2‐. The formation constant for the reaction is shown below. (Be sure to account for the speciation of EDTA at the lake pH.) Ca2+ + EDTA4‐ <==> CaEDTA2‐ log K = 12.4 SOLUTION: There are 4 important unknowns: Ca2+, CaEDTA2‐, H2EDTA2‐ and HEDTA3‐, but we will also need to consider EDTA4‐. The four equations are: 1. CT‐Ca = [Ca2+] + [CaEDTA2‐] = 5x10‐4 2. CT‐EDTA = [H2EDTA2‐] + [HEDTA3‐] + [CaEDTA2‐] = 10‐5 3. Ka3 = [HEDTA3‐][H+]/[H2EDTA2‐] 4. Β = [CaEDTA2‐]/[Ca2+][EDTA4‐] 5. Ka4 = [EDTA4‐][H+]/[HEDTA3‐] Rearranging: [Ca2+] = CT‐Ca/(1+β[EDTA4‐] [CaEDTA2‐] = CT‐Ca(1‐1/(1+β[EDTA4‐])) [HEDTA3‐] = [EDTA4‐][H+]/Ka4 [H2EDTA2‐] = ([EDTA4‐][H+]/Ka4)*[H+]/Ka3 Subsituting into the EDTA mass balance: CT‐EDTA = [EDTA4‐]([H+]2/Ka3Ka4 + [H+]/Ka4) + CT‐Ca (1‐1/(1+β[EDTA4‐])) Solving iteratively, the EDTA conc. of 10‐14.09 (8.15x10‐15) may be calculated. This allows the conc. of CaEDTA to be calculated as 1x10‐5 M or 2% of total Ca2+. C) To examine the effect of EDTA on trace metal speciation, we must first determine the speciation in the absence of EDTA. Consider a lake with a total Cu2+ concentration of 10‐7 M at pH 8.0 and with an alkalinity of 1 mM. The major inorganic complex affecting Cu2+ is the carbonate complex: Cu2+ + CO32‐ <==> CuCO3o log K = 6.7 Calculate the concentration of free Cu2+ in this lake (i.e., consider only the carbonate complex). SOLUTION: The approach is similar to that above, but because carbonate is present in excess of Cu, we can assume that the free [CO32‐] concentration is the same as the total carbonate ion concentration that may be calculated from the alkalinity. Alk = 10‐3 M = [HCO3‐] + 2[CO32‐] + [OH‐] – [H+] Alk = [ HCO3− ] + 2[CO32− ] + [OH − ] − [ H + ]
Alk =
[CO32− ][ H + ]
K
+ 2[CO32− ] + w+ − [ H + ]
Ka2
[H ]
Kw
[H + ]
[H + ]
2+
Ka2
Alk + [ H + ] −
2−
3
[CO ] =
[CO32− ] = 4.96 x10−6 = 10−5.30
Now we can use the Cu mass balance equation to find the fraction free and bound Cu: CT = [Cu2+] + [CuCO3o] = [Cu2+](1+ β[CO32‐]) [Cu2+] = CT/(1+ β[CO32‐]) = 10‐7/(1+106.7∙10‐5.3) = 1x10‐14 D) Now calculate the concentration of free Cu2+ in the lake if the total concentration of EDTA in the lake were 10‐5 M. The binding constant for Cu(II)l‐EDTA is given in Appendix D of the text. SOLUTION: We can use information from the parts above to arrive at an answer. We might first try the assumption that the EDTA will bind essentially all of the Cu CT‐Cu = [Cu2+] + [CuCO3o] + [CuEDTA] ≈ [CuEDTA] = 10‐7 Because this is only 1% of the total EDTA, it might not affect the distribution of EDTA species very much so we could use the concentration of free [EDTA4‐] (= 10‐14.09) to estimate the free Cu: Β = [CuEDTA2‐]/[Cu2+][EDTA4‐] [Cu2+] = [CuEDTA2‐]/β[EDTA4‐] = 10‐7/(1018.78)(10‐14.09) = 2.04x10‐12 We know that is not correct because it is higher than the value we calculated for the case of only Carbonate binding. We must, therefore include both complexes of Cu in the Cu mass balance equation: CT‐Cu = [Cu2+] + [CuCO3o] + [CuEDTA] = [Cu2+](1 + βCO3[CO32‐] + βEDTA[EDTA4‐] We could either use the free EDTA calculated above or calculate it from the assumption that total EDTA ≈ [HEDTA3‐] [EDTA4‐] = Ka4[HEDTA3‐]/[H+] ≈ Ka4CT‐EDTA/[H+] Now, free Cu may be calculated as: CT −Cu
[Cu 2+ ] =
C
1 + βCO 3 [CO32− ] + β EDTA K a 4 T − EDTA
[H + ]
This approach gives an estimate of free Cu2+ of 2.2x10‐18M.