Question

Calculate pCu2+ at each of the following points in the titration of 50.00 mL of 0.001 00 M Cu2+ with 0.00100 M EDTA at pH 11.00 in a solution with [NH3] fixed at 1.00 M:

(a) 0 mL (c) 45.00 mL (e) 55.00 mL (b) 1.00 mL (d) 50.00 mL

Please show all steps to solution; thank you.

Answer 1

Answer 2

EDTA is a tetrabasic acid that has six potential sites

for bonding with a metal ion, four carboxyl groups

and the two amino groups , each of the latter with

unshared pair of electrons. Thus EDTA is a

hexadentate ligand.

The dissociation constants for the acidic groups in

EDTA are K1= 1.02x10-2, K2= 2.14x10-3,

K3= 6.92x10-7 and K4 = 5.50x10-11.

%uF06E The various EDTA species are often

abbreviated as H4Y, H3Y-, H2Y2-, HY3- and Y4-.

The fully unprotonated form Y4- is dominant only in very

basic solutions (pH>10).here pH=11

Cu2+ can also be titrated in alkaline medium

(pH 10) using ammoniacal buffer and Eriochrome

black T as indicator.

EDTA is not a selective reagent (G.R.) as it can complex

with almost all metal ions (cations)except alkali metals in a

ratio 1:1 regardless of the charge on the cation.

e.g. Mg2+ + H2Y2- -------- MgY2-

Al3+ + H2Y2- ------------- AlY-

so for Cu2+ it reacts with 1:2 ratio

Answer 3

EDTA is a tetrabasic acid that has six potential sites

for bonding with a metal ion, four carboxyl groups

and the two amino groups , each of the latter with

unshared pair of electrons. Thus EDTA is a

hexadentate ligand.

The dissociation constants for the acidic groups in

EDTA are K1= 1.02x10-2, K2= 2.14x10-3,

K3= 6.92x10-7 and K4 = 5.50x10-11.

%uF06E The various EDTA species are often

abbreviated as H4Y, H3Y-, H2Y2-, HY3- and Y4-.

The fully unprotonated form Y4- is dominant only in very

basic solutions (pH>10).here pH=11

Cu2+ can also be titrated in alkaline medium

(pH 10) using ammoniacal buffer and Eriochrome

black T as indicator.

EDTA is not a selective reagent (G.R.) as it can complex

with almost all metal ions (cations)except alkali metals in a

ratio 1:1 regardless of the charge on the cation.

e.g. Mg2+ + H2Y2- -------- MgY2-

Al3+ + H2Y2- ------------- AlY-

so for Cu2+ it reacts with 1:2 ratio

Answer 4

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