Search... n/ibiscms/mod/ibis/view.phpid 4988480 linglearning.com G The EDTA solu
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Search... n/ibiscms/mod/ibis/view.phpid 4988480 linglearning.com G The EDTA solution is buffered a. saplinglearning.com Jump to.. Map Sapling Learning Calculate pva at each of the following points in the titration of 24.01 mL of 0.0530 M EDTA with 0.0265 M VCh The EDTA solution is buffered at a pH of 10.00. The fraction of free EDTA in the Y form (ay) can be found here. The formation constant for the V-EDTA complex is given by log Ky 12.7 a) 4.802 ml 10 48.02 mL Number Number b) 19.21 mL 9)48.12 ml Number Number c) 38.42 ml h) 52.82 mL Number Number Scroll down to vievw the entire question. d) 47.06 mL 57.62 mL Hint O Prerous"oCheck AnswerNext t ExitExplanation / Answer
For EDTA,
alpha[Y4-] at pH 10 = 0.36
given log Kf = 12.7
Kf = 5.0 x 10^12
Kf' = Kf x alpha[Y4-] = 5.0 x 10^12 x 0.36 = 1.8 x 10^12
a) 19.21 ml VCl2 added
moles EDTA = 0.0530 M x 24.01 ml = 1.27253 mmol
moles VCl2 = 0.0265 M x 4.802 ml = 0.127253 mmol
[VEDTA^2-] formed = 0.127253 mmol/(24.01 + 4.802) ml = 4.42 x 10^-3 M
excess [EDTA] = (1.27253 - 0.127253 mmol/(24.01 + 4.802) ml = 0.04 M
[V2+] = [VEDTA^2-]/Kf[EDTA]
= 4.42 x 10^-3/1.8 x 10^12 x 0.04
= 6.18 x 10^-14 M
pV2+ = -log[V2+] = 13.21
b) 19.21 ml VCl2 added
moles EDTA = 0.0530 M x 24.01 ml = 1.27253 mmol
moles VCl2 = 0.0265 M x 19.21 ml = 0.509065 mmol
[VEDTA^2-] formed = 0.509065 mmol/(24.01 + 19.21) ml = 0.012 M
excess [EDTA] = (1.27253 - 0.509065 mmol/(24.01 + 19.21) ml = 0.018 M
[V2+] = [VEDTA^2-]/Kf[EDTA]
= 0.012/1.8 x 10^12 x 0.018
= 3.8 x 10^-13 M
pV2+ = -log[V2+] = 12.43
c) 38.42 ml VCl2 added
moles EDTA = 0.0530 M x 24.01 ml = 1.27253 mmol
moles VCl2 = 0.0265 M x 38.42 ml = 1.01813 mmol
[VEDTA^2-] formed = 1.01813 mmol/(24.01 + 38.42) ml = 0.01631 M
excess [EDTA] = (1.27253 - 1.01813) mmol/(24.01 + 38.42) ml = 4.1 x 10^-3 M
[V2+] = [VEDTA^2-]/Kf[EDTA]
= 0.01631/1.8 x 10^12 x 4.1 x 10^-3
= 2.22 x 10^-12 M
pV2+ = -log[V2+] = 11.65
d) 47.06 ml VCl2 added
moles EDTA = 0.0530 M x 24.01 ml = 1.27253 mmol
moles VCl2 = 0.0265 M x 48.12 ml = 1.24709 mmol
[VEDTA^2-] formed = 1.24709 mmol/(24.01 + 47.06) ml = 0.01755 M
excess [EDTA] = (1.27253 - 1.24709) mmol/(24.01 + 47.06) ml = 3.6 x 10^-4 M
Kf' = [VEDTA^2-]/[V2+][EDTA]
[V2+] = 0.01755/1.8 x 10^12 x 3.6 x 10^-4
= 2.708 x 10^-11 M
pV2+ = -log[V2+] = 10.57
f) 48.02 ml VCl2 added
moles EDTA = 0.0530 M x 24.01 ml = 1.27253 mmol
moles VCl2 = 0.0265 M x 48.12 ml = 1.27253 mmol
Equivalence point
[VEDTA^2-] formed = 1.27253 mmol/(24.01 + 48.02) ml = 0.018 M
V2+ + EDTA <====> VEDTA^2-
I - - 0.018
C +x +x -x
E x x 0.018-x
let x be a small change then,
Kf' = [VEDTA^2-]/[V2+][EDTA]
1.8 x 10^12 = 0.018/x2
x = [V2+] = 1 x 10^-7 M
pV2+ = -log[V2+] = 7.0
g) 48.12 ml VCl2 added
moles EDTA = 0.0530 M x 24.01 ml = 1.27253 mmol
moles VCl2 = 0.0265 M x 48.12 ml = 1.27518 mmol
excess [VCl2] = [V2+] = (1.27518 - 1.27253) mmol/(24.01 + 48.12) ml = 3.7 x 10^-5 M
pV2+ = -log[V2+] = 4.43
h) 52.82 ml VCl2 added
moles EDTA = 0.0530 M x 24.01 ml = 1.27253 mmol
moles VCl2 = 0.0265 M x 52.82 ml = 1.39973 mmol
excess [VCl2] = [V2+] = (1.39973 - 1.27253) mmol/(24.01 + 52.82) ml = 1.6 x 10^-3 M
pV2+ = -log[V2+] = 2.78
i) 57.62 ml VCl2 added
moles EDTA = 0.0530 M x 24.01 ml = 1.27253 mmol
moles VCl2 = 0.0265 M x 57.62 ml = 1.52693 mmol
excess [VCl2] = [V2+] = (1.52693 - 1.27253) mmol/(24.01 + 57.62) ml = 3.1 x 10^-3 M
pV2+ = -log[V2+] = 2.51
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