calculate pMn2+ at each of the following points in the titration of 22.56ml of 0
ID: 894937 • Letter: C
Question
calculate pMn2+ at each of the following points in the titration of 22.56ml of 0.0538 M EDTA with 0.0269M MnCl2. The EDTA solution is buffered at a pH of 10.00. the formation constant for the Mn2+-EDTA complex is given by log K1=13.89.
a. 4.512 mL b. 18.05mL c. 36.10mL d. 44.22mL e. 45.03mL f 45.12mL g.45.21mL h 49.64mL f. 54.14 mL
Values of Y4-for EDTA at 25°C and =0.10M
pH Y4- 0 1.3 x 10-23 1 1.4 x 10-18 2 2.6 x 10-14 3 2.1 x 10-11 4 3.0 x 10-9 5 2.9 x 10-7 6 1.8 x 10-5 7 3.8 x 10-4 8 4.2 x 10-3 9 0.041 10 0.30 11 0.81 12 0.98 13 1.00 14 1.00Explanation / Answer
pMn2+ calculations.
a) when 4.512 ml of 0.0269 M was added
moles of EDTA = molarity x volume = 0.0538 x 22.56 = 1.214 mmols
moles of MnCl2 = 0.0269 x 4.512 = 0.121 mmol
molarity of [MnY^2-] = 0.121/27.072 = 4.47 x 10^-3 M
So, 0.121 mmols of MnCl2 will reacts with EDTA
remaining moles of EDTA = 1.093 mmol
Total volume of solution = 22.56 + 4.512 = 27.072
molarity of [EDTA] = 1.093/27.072 = 0.0404 M
Kf.[Y^4-] = [MnY^2-]/[Mn2+][EDTA]
Feed values,
7.76 x 10^13 x 0.30 = 4.47 x 10^-3/[Mn2+](0.0404) = 2.33 x 10^13
[Mn2+] = 4.75 x 10^-15 M
pMn2+ = log[Mn2+] = 14.32
d) 44.22 ml of MnCl2 added
moles of EDTA = molarity x volume = 0.0538 x 22.56 = 1.214
moles of MnCl2 = 0.0269 x 44.22 = 1.19 mmol
molarity of [MnY^2-] = 0.121/27.072 = 0.018 M
So, 1.19 mmols of MnCl2 will reacts with EDTA
remaining moles of EDTA = 0.0245 mmol
Total volume of solution = 44.22 + 22.56 = 66.78 ml
molarity of EDTA = 0.0245/66.78 = 3.69 x 10^-4 M
Kf.[Y^4-] = [MnY^2-]/[Mn2+][EDTA]
Feed values,
2.33 x 10^13 = 0.018/[Mn2+](3.69 x 10^-4)
[Mn2+] = 2.09 x 10^-12 M
pMn2+ = log[Mn2+] = 11.68
f) 45.12 ml of MnCl2 added
moles of EDTA = molarity x volume = 0.0538 x 22.56 = 1.214
moles of MnCl2 = 0.0269 x 45.12 = 1.214 mmol
molarity of [MnY^2-] = 1.214/67.68 = 0.018 M
So, we have only MnY^2- at this point
let x amount of complex has dissociated then,
Kf.[Y^4-] = [MnY^2-]/[Mn2+][EDTA]
Feed values,
2.33 x 10^13 = 0.018-x/(x)(x)
with x be a small amount,
2.33 x 10^13 = 0.018/x^2
x = [Mn2+] = 2.78 x 10^-8 M
pMn2+ = log[Mn2+] = 7.56
h) 49.64 ml of MnCl2 added
moles of EDTA = molarity x volume = 0.0538 x 22.56 = 1.214
moles of MnCl2 = 0.0269 x 49.64 = 1.34 mmol
excess Mn2+ = 0.12 mmol
free [Mn2+] = 0.12/(22.56+49.64) = 1.66 x 10^-3 M
pMn2+ = -log(1.66 x 10^-3) = 2.78
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