Al3+ + Y4- --> AlY-
Kf = 1016.4
= 2.51 x 1016
Kf’ = Kf * αY4-
Kf’ = 2.51 x 1016 * 0.98 = 2.46 x 1016
Kf’ = [AlY-] / ([Y4-] [Al3+])
Moles of Al3+ added = 100 mM / 1000 mM/M * 50 mL = 5 mmoles
a)
[Al3+] = 100 x 10-3 M
= 0.1 M
pAl3+ = - log([Al3+])
= 1
b)
Total Volume = 50 + 25 = 75 mL
Moles of EDTA added = 100 mM / 1000 mM/M * 25 mL = 2.5 mmoles
Al3+ + Y4- --> AlY-
(5 – x) + (2.5 – x) --> x
Divide by volume to get concentration.
Kf’ = [AlY-] / ([Y4-] [Al3+])
Kf’ = x * 75 / ((5 – x) (2.5 – x)) = 2.46 x 1016
x = 2.5
[Al3+] = (5 – x)/75 = 0.033
pAl3+ = 1.48
c)
Total Volume = 50 + 50 = 100 mL
Moles of EDTA added = 100 mM / 1000 mM/M * 50 mL = 5 mmoles
Al3+ + Y4- --> AlY-
(5 – x) + (5 – x) --> x
Divide by volume to get concentration.
Kf’ = [AlY-] / ([Y4-] [Al3+])
Kf’ = x * 100 / ((5 – x) (5 – x)) = 2.46 x 1016
x = 4.99999986
[Al3+] = (5 – x)/100 = 1.4 x 10-9
pAl3+ = 8.85
d)
Total Volume = 50 + 100 = 150 mL
Moles of EDTA added = 100 mM / 1000 mM/M * 100 mL = 10 mmoles
Al3+ + Y4- --> AlY-
(5 – x) + (10 – x) --> x
Divide by volume to get concentration.
Kf’ = [AlY-] / ([Y4-] [Al3+])
Kf’ = x * 150 / ((5 – x) (10 – x)) = 2.46 x 1016
x = 4.99999994
[Al3+] = (5 – x)/150 = 4 x 10-10
pAl3+ = 9.4
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