Very tricky question. I need help working it out properly and finding the answer

Question

Very tricky question. I need help working it out properly and finding the answer. Thank you

The equilibrium constant for the following reaction is 1.0 × 1023. Cr3+ (aq) + H2 EDTA2-(aq) ? CrEDTA-(aq) + 2H+ (aq) o,c-CHN-CH2-CH2-N O2C-CH2 CH2-CO2 EDTA-=-O2C-CH2 Ethylenediaminetetraacetate EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt Na2H2 EDTA, are used to treat heavy metal poisoning. Calculate Cr3at equilibrium in a solution originally 0.0090 Min Cr+ and 0.065 Min H2 EDTA2 and buffered at pH- 6.00. as

Explanation / Answer

The reaction between chroium ion and EDTA is :

Cr+3(aq) + H?EDTA-2(aq) == CrEDTA?(aq) + 2H+(aq)

Given that

Kc = 1.0 × 1023
pH of solution is 6

We know that

pH = -log[H+]

so [H+]= 10-6
The solution is buffered at pH = 6.00
[H?] thorough the process = 1 × 10?? M

The reaction starts with the following Initial concentrations
Cr+3(aq) = 0.0010 M
H?EDTA-2(aq) = 0.050 M
CrEDTA?(aq) = 0 M
As per balanced equation one mole of chromium ion will complex with one mole of EDTA , so there is excess of EDTA as comapred to Chromium ion, hence the limiting reagent is chromium ion.

At the end of reaction there will be no chromium ion left, as the reaction may go to completion as the Kc is very very high
We can say that the equilibrium concentration of
H?EDTA-2(aq) = 0.050 + 0.001 = 0.051 M

CrEDTA?(aq) = initial concentration of chromium ion = 0.001 M
Putting these in equilibrium expression as

Kc = [CrEDTA?] [H+]2 / {[Cr+3] [H?EDTA-2]}
1023 = 0.001 × (10-6)2/ [[Cr+3] × 0.051]

Hence
[Cr+3] = 0.001 × (10-6)2/ [0.051 × (1023)] = 10-14 / 0.051 X 1023

[Cr+3] = 1.96 X 10-36 M

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