CHE 230S ENVIRONMENTAL CHEMISTRY
PROBLEM SET 6 – Full Solutions
Easier problems
1) Given below is a partially completed pE/pH diagram for arsenic species.
H3AsO4
H2AsO4HAsO42H3AsO3
a) Write a chemical reaction that describes the line shown for H3AsO3 and H2AsO4AsO4-(aq)+2e+3H+(aq)
AsO3(aq)+H2O(l)
b) Based on the figure, give the equation relating pE and pH, of the line shown for H 3AsO3 and
H2AsO4-.
From the diagram, when pH = 0, pε ≈ 11
Since the “e” to “H+” stoichiometry in the reaction is 2 to 3, the slope is 3/2
Therefore the equation is pε ≈ 11 – 3/2 pH
c) Estimate the values of εo and ∆Gorxn for the reaction in part a based on the equation form
part b. (0.65V, -125.4kJ/mol)
From the reaction in part a
log K = 2 εo/ 0.0591 = 2 pεo + 3 pH
From part b
pε ≈ 11 – 3/2 pH
2 pεo + 3 pH =22
Therefore εo/ 0.0591 ≈ 22/2 = 11 and εo ≈ 0.65V
∆Gorxn = -nF εo = -2 (96485) (0.65) = -125.4 kJ/mol
d) Based on the figure, which form of arsenic would you expect to predominate in i) acid mine
drainage (oxidizing with pH <3), ii) sea water (oxidizing with pH = 8.5) and iii) ground water
(very reducing)? What are the implications in terms of arsenic toxicity?
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) because of low pH (pH<3).
The seawater species will also be of higher oxidation state but likely deprotonated (H2AsO4-)
because of higher pH (pH>8.5). In ground water it will be in the reduced state (As (III)) which is
more toxic.
2) Cadmium has recently become a concern due to its presence in children’s jewellery
(http://www.cbc.ca/health/story/2010/01/11/consumer-cadmium-childrens-jewelry.html). The
concern is that Cadmium could leach out of this jewellery and thereby enter the body. The
solubility of cadmium is influenced by complex formation. Given below are the logarithms of
the stepwise formation constants of various ligands with Cd 2+ (values expressed in M):
Ligand
K1
K2
OH-
6.08
2.62
Cl-
2.00
0.7
a) Name the complex that predominates for pH = 10, pCl = 1 (Hydroxycadmium(II))
If pH=10, pOH=4
Total Cd = [Cd2+]+[Cd(OH)+]+[Cd(OH)2]+[CdCl+]+[CdCl2]
= [Cd2+] (1+106.08-4+ 106.08+2.62-8+102-1+102+0.7-2)
=[Cd2+] (1+102.08+ 100.7+101+100.7)
Therefore Cd(OH)+ predominates
b) Given that the solubility product of Cd(OH)2(s) is 2*10-14, calculate the molar solubility of
cadmium hydroxide in water with pCl = 1 and pH fixed at 9 (5.6x10-3 M).
[Cd2+][OH-]2=Ksp=2*10-14
If pH=9, pOH=5
[Cd2+]=2*10-14/(10-5*2)= 2*10-4M
Total Cd = [Cd2+]+[Cd(OH)+]+[Cd(OH)2]+[CdCl+]+[CdCl2]
= [Cd2+] (1+106.08-5+ 106.08+2.62-10+102-1+102+0.7-2*1)
=2*10-4 (1+101.08+ 10-1.3+101+100.7)
=2*10-4 (28.1)
= 5.62x10-3 M
3) Large quantities of arsenic enter the environment through the combustion of coal and
leaching from mine tailings. Much of this arsenic reaches natural water systems. The toxicity of
arsenic depends upon its chemical speciation, its oxidation state in particular: As(III) is much
more toxic than As (V). Use the redox reactions given below to answer the following questions:
Data:
O2(g) + 4H+ + 4e- -----> 2H2O
2H2O + 2e- -----> H2(g) + 2OHH3AsO4 + 2H+ + 2e- -----> H3AsO3 + H2 O
εo = 1.23 V
εo = -0.83 V
εo = 0.58 V
a) What is the pH of water, at 25oC in equilibrium with 1 atm air, with a pE of 15 (5.62)
O2(g) + 4H+ + 4e- -----> 2H2O εo = 1.23 V
The Nernst equation for this reaction is
E = E0 – 0.05916/4 * log(K)
And K is given by
K=1/(PO2*[H+]4*[e-]4)
Or
logK= -log PO2-4log[H+]-4log[e-]
Rearranging the Nernst equation gives
log(K) = 4 ( E0 – E) / 0.05916 = 4 ( E0) / 0.05916
since E = 10-15V << E0 = 1.23V
Combining the two:
4 ( E0) / 0.05916 = -log PO2-4log[H+]-4log[e-]
( E0) / 0.05916 = -log (PO2)/4 + pH + pE
( E0) / 0.05916 = -log (PO2)/4 + pH + pE
pH=E0/0.5916 + log(PO2)/4 – pE
Since PO2=0.21 atm,
pH=1.23/0.0591+log(0.21)/4-15=5.62
b) Which oxidation state of As predominates in the water from part a (As(V))
V
III
H3AsO4 + 2H+ + 2e- -----> H3AsO3 + H2 O
εo = 0.58 V
+ 2 - 2
o
logK=log([H3AsO3]/ [H3AsO4][H ] [e ] )= 2* ε /0.0591
logK=log([H3AsO3]/ [H3AsO4])+2pH+2pE= (2*0.58)/0.0591
logK=log([H3AsO3]/ [H3AsO4])=( (2*0.58)/0.0591)-(2*5.63)-(2*15)= -21.6
Therefore, [H3AsO3]<< [H3AsO4] and As(V) predominates
More challenging problems
4) Lead in drinking water is an ongoing concern (http://www.toronto.ca/health/lead/drinking_
water.htm). Lead forms highly insoluble sulphide and hydroxide precipitates which might lead
one to believe that the lead concentration in water would be very low. However, lead solubility
is greatly influenced by complex formation.
For PbS Ksp = [Pb2+][S2-] = 3.4*10-28. Ka values for H2S and HS- are 9.5*10-8 and 1.0*10-19
respectively. The log stability constants are 7.82, 3.06 and 3.06. Calculate the solubility of PbS(s)
in water at a) pH 5 and b) pH 9. Neglect any loss of H2S due to volatilization. (1.9*10-6 M,
5.0*10-8 M)
c) Given that Ksp = 2.5*10-16 for Pb(OH)2(s), would lead hydroxide precipitate be formed at
either pH ? (no)
a) pH 5 ,
[Pb2+][S2-] = 3.4*10-28
Let “x” equal to the total concentration of lead species or total concentration of sulphur species
x= [Pb2+]+[Pb(OH)+)+[Pb(OH)2]+[Pb(OH)3-]
= [Pb2+] (1+k1[OH]-+β2[OH]2 +β3[OH]3)
= [Pb2+] (1+107.82[OH]+ 1010.88[OH]2+1013.94[OH]3)
x= [S2-]+[HS-]+[H2S]
= [S2-] (1+1019[H+] + 1019+7.02[H+]2)
S
Ka1=9.5 * 10-8
Ka2 =10-19
At pH=5 (pOH=9),
[Pb(OH)2] and [Pb(OH)3-] are very small (1010.88-(2*9) and 1013.94-(3*9)) and can be disregarded.
Therefore,
x= [Pb2+] (1+107.82-9) = [Pb2+] (1.07)
x= [S2-] (1+1019-5 + 1026-(2*5)) = [S2-](1016]
Therefore, [Pb2+][S2-] = (x/1.07) * (x/1016) = 3.4*10-28
x2 = 3.64*10-12
x=1.91*10-6M
b) pH 9,
At pH=9 (pOH=5),
x= [Pb2+] (1+107.82-5+ 1010.88-(2*5)+1013.94-(3*5))
= [Pb2+] (668.6)
x= [S2-] (1+1019-9 + 1026-(2*9))
= [S2-] (1.1*1010)
Therefore, [Pb2+][S2-] = (x/668.6) * (x/1.1*1010) = 3.4*10-28
x2 = 2.5*10-15
x=5*10-8M
c) Given that Ksp = 2.5*10-16 for Pb(OH)2(s), would lead hydroxide precipitate be formed at
either pH ?
At pH=5, Q = [Pb2+][OH-]2 = 1.9*10-6/1.07 * (10-9*2) = 1.79*10-24
Q<<Ksp, therefore lead hydroxide precipitate would not be formed
At pH=9, Q = [Pb2+][OH-]2 = 5*10-8/668.6 * (10-5*2) = 7.5*10-21
Q<<Ksp, therefore lead hydroxide precipitate would not be formed
5) Nitrogen is usually present in water in forms ranging from NH4+ to NO3- depending on the
redox potential of the system. Based on the reactions
2NO3- (aq) + 12H+ (aq) + 10 e- ↔ N2(g) + 6 H2O(l) εo = 1.24 V
NO3- (aq) + 2H+ (aq) + 2 e- ↔ NO2- (aq) + H2O(l) εo = 0.836 V
NO2- (aq) + 8H+ (aq) + 6 e- ↔ NH4+(aq) + 2 H2O(l) εo = 0.89 V
a)
Calculate the pE at which the concentration of NO3-(aq) is 100 times that of NO2-(aq) for
water with a pH of 6. (9.1)
NO3- (aq) + 2H+ (aq) + 2 e- ↔ NO2- (aq) + H2O(l) εo = 0.836 V
logK=2*εo /0.0591=log([NO2-]/([NO3-][H+]2[e-]2)
log([NO2-]/([NO3-])=(2*0.836)/0.0591)-2pH-2pE = 28.29 -2pH-2pE = log(1/100)
Since pH=6,
log(1/100)= 28.29 -(2*6)-2pE
pE= 9.1
b)
Calculate the pE at which NO3-(aq) and NH4+(aq) have equal concentrations in water with a
pH of 8. (4.8)
NO2- (aq) + 8H+ (aq) + 6 e- ↔ NH4+(aq) + 2 H2O(l)
Log([NH4]/([NO2-][H+]8[e-]6)=(0.89*6)/0.0591=90.35
Log([NH4]/([NO2-])=90.35-8pH-6pE
Combine with answer form part a:
Log([NH4]/([NO3-])= Log([NH4]/([NO2-])+ Log([NO2]/([NO3-])= 90.35-8pH-6pE+28.29-2pH-2pE
Therefore, 118.64-10pH-8pE=0
pE=(118.64-10pH)/8=(118.64-(10*8))/8=4.8
c)
Calculate the concentration of NH4+(aq) in water with a pH of 4 and pE = 0, in equilibrium
with air. PN2 is 0.8atm. (0.005 M)
2NO3- (aq) + 12H+ (aq) + 10 e- ↔ N2(g) + 6 H2O(l) εo = 1.24 V
Log(PN2/([NO3-]2[H+]12[e-]10) =Log(PN2/[NO3-]2)+12pH+10pE = (1.24*10)/0.0591 = 209.81
At pH of 4 and pE = 0,
Log(0.8/[ NO3-]2)= 209.81-(12*4)= 161.8
[NO3-]= (0.8*10 -161.8)0.5 =1.1*10-81M
From b, Log([NH4]/([NO3-])=118.64-10pH-8pE
Log([NH4+]/1.1*10-81M)=118.65-(10*4)-0=78.65
Therefore, [NH4+]=(1078.65)*(1.1*10-81) = 0.0049M