Question

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Quiz 9 Name: (35 pts) You will titrate 50.00 mL of 100.0 mM AlP with 100.0 mM EDTA solution, all at pH 12.00 What is the pA when add 50 mL of 100.0 mM EDTA solution? for Al binding to Y* Kr 10164 atpH 12.00 αY4. = 0.98 What is the pAI when add a) 0 mL of 100.0 mM EDTA b) 25.00 mL of 100.0 mM EDTA c) 50.00 mL of 100.0 mM EDTA d) 100.00 mL of 100.0 mM EDTA

Answer 1

Al³⁺ + Y^4- --> AlY^-

Kf = 10^16.4

= 2.51 x 10¹⁶

Kf’ = Kf * α_Y4-

Kf’ = 2.51 x 10¹⁶ * 0.98 = 2.46 x 10¹⁶

Kf’ = [AlY^-] / ([Y^4-] [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⁺ + Y^4- --> AlY^-

(5 – x) + (2.5 – x) --> x

Divide by volume to get concentration.

Kf’ = [AlY^-] / ([Y^4-] [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⁺ + Y^4- --> AlY^-

(5 – x) + (5 – x) --> x

Divide by volume to get concentration.

Kf’ = [AlY^-] / ([Y^4-] [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⁺ + Y^4- --> AlY^-

(5 – x) + (10 – x) --> x

Divide by volume to get concentration.

Kf’ = [AlY^-] / ([Y^4-] [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