1. Create a logC-pH diagram for the speciation of lead ion (Pb2+) and its hydroxy complexes. Assume Pb2,T = 0.01 mM. Use the beta values from Table 9.3. Do not consider Pb(OH)2 precipitation. a. At pH...

1.
    pH
    [OH-]
    [Pb2+]
    Log [Pb2+]
    log [Pb(OH)+1]
    Log [Pb(OH)2]
    Log [Pb(OH)3]
    Log [Pb(OH)4]
    1.00
    1E-13
    1E-05
    -5.00
    -11.60
    -20.10
    -30.10
    -40.71
    2.00
    1E-12
    9.99997E-06
    -5.00
    -10.60
    -18.10
    -27.10
    -36.71
    3.00
    1E-11
    9.99975E-06
    -5.00
    -9.60
    -16.10
    -24.10
    -32.71
    4.00
    1E-10
    9.99749E-06
    -5.00
    -8.60
    -14.10
    -21.10
    -28.71
    5.00
    0.000000001
    9.97494E-06
    -5.00
    -7.60
    -12.10
    -18.10
    -24.71
    6.00
    0.00000001
    9.75489E-06
    -5.01
    -6.61
    -10.11
    -15.11
    -20.72
    7.00
    0.0000001
    7.98733E-06
    -5.10
    -5.70
    -8.20
    -12.20
    -16.81
    8.00
    0.000001
    2.78443E-06
    -5.56
    -5.16
    -6.66
    -9.66
    -13.27
    9.00
    0.00001
    2.92896E-07
    -6.53
    -5.13
    -5.63
    -7.63
    -10.24
    10.00
    0.0001
    8.86607E-09
    -8.05
    -5.65
    -5.15
    -6.15
    -7.76
    11.00
    0.001
    5.52999E-11
    -10.26
    -6.86
    -5.36
    -5.36
    -5.97
    12.00
    0.01
    3.54689E-14
    -13.45
    -9.05
    -6.55
    -5.55
    -5.16
    13.00
    0.1
    4.93684E-18
    -17.31
    -11.91
    -8.41
    -6.41
    -5.02
    14.00
    1
    5.11951E-22
    -21.29
    -14.89
    -10.39
    -7.39
    -5.00
a) It is clearly revealed from the graph that at pH 10 Pb(OH)2 concentration is dominating only.
b) At pH 7, [Pb 2+] = 10 -5.097
[Pb(OH+)] = B1x[Pb 2+]x[OH-]
[Pb(OH)2] = B2x[Pb 2+]x[)OH-]2
[Pb(OH+)]/[Pb(OH)2] = B1/B2*[OH-]
= 106.4/( 1010.9x 10-7) = 101.23 = 16.98
2.
If concentration of Cu2+ is fixed, equations (1) to (4) become straight lines on the Logarithmic Concentration Diagram.
Equation (1) becomes
a. [Cu(OH+) / [Cu2+] = B1*[Cu2+]*[OH-]/[Cu2+] = 106.5*10-7 = 0.316
b. [NH4+]/[NH3] = K1. [H+] = 103.16.10-7 = 0.000144
c. No there is no hydroxo complex of Cu present in greater than [Cu(OH+)]
d. from Table 9.4
log K1 = 3.99
Kb = 1.86x 10-5
[NH3] = [NH4+][OH-]/ (1.86x 10-5)
    = 10 -3x 10-7) / (1.86x 10-5) = (5.38x 10-6)=
Similarly,
So from the calculation it reveals that log[Cu(NH3)2+] is highest so, this complex concentration is maximum.
It can be prove in other ways also,
3. B1 = 8.8
B2 = 18.4
Now the concentration are equal
So, B1*[Hg2+]*[NH3] = B2*[Hg2+]*[NH3]3
[NH3] = (B1/B2)0.5 = (108.8/1018.4)0.5 = 1.86
So, Partial pressure of NH3 = KH*[NH3]
=0.0171x1.86 atm = 0.032...