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When there is $\ce{N^m+}$ and $\ce{M^n+}$ metal ions in a solution with respective m and n charge,

$\ce{N^{m+}, M^{n+}}$ reacts with EDTA($\ce{Y}$) as follows:

$$\ce{N^m+ + Y^4- <=> NY^{m-4}}$$

$$\ce{M^n+ + Y^4- <=> MY^{n-4}}$$

Formation constant for each chemical reaction,

$$\ce{K_{NY^{m-4}}} = \frac{\ce{[NY^{m-4}]}}{\ce{[N^{m+}][Y^{4-}]}}$$

$$\ce{K_{MY^{n-4}}} = \frac{\ce{[MY^{n-4}]}}{\ce{[M^{n+}][Y^{4-}]}}$$

where as $\ce{N^m+}$ has the higher value for formation constent KNYm-4 > KMYn-4

Fraction of each ion exsists ${α}$, $\ce{[N_{total}][M_{total}][Y_{total}]}$ are the total concentraion of the ion in any form.

$\ce{α_{Y^{4-}}}$ = $\frac{\ce{[Y^4-]}}{\ce{[Y_{total}]}}$ $\ce{α_{N^{m+}}}$ = $\frac{\ce{[N^m+]}}{\ce{[N_{total}]}}$ $\ce{α_{M^{n+}}}$ = $\frac{\ce{[M^n+]}}{\ce{[M_{total}]}}$

After $\ce{N^m+}$ has reacted with EDTA,the remaining amount EDTA is used for $\ce{M^n+}$

$$\ce{[Y^{'}_{total}] = [Y_{total}] + [NY^{m-4}]}$$

Where as $\ce{α_Y^{'}}$ is,

$$\ce{α_{Y^{'}}} = \frac{\ce{[Y_{total}]}}{\ce{[Y^{'}_{total}]}}$$

These are all the definition constants but I have no clue how this equation came

$$\ce{K''_{MY^{n-4}} = K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}} - α_Y^{'}}$$

To perform a selective complexometric titration this condition need to be satisfied,

$$\ce{K''_{MY^{n-4}} = K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}} - α_Y^{'} \geq 10^7}$$

where αY' is given below,

$$\ce{α_{Y'}} = \frac{1}{\ce{(1 + K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}])}}$$ Using EDTA as for an example Y4-

How did this equation came with the assumptions (The simplified equation)

$$\ce{K''_{MY^{n-4}}} = \frac{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}}{K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}]}$$

One of the assumptions would be

$$\ce{ K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] \gg 1}$$

Addition of one can be neglected.

What are the assumption do need to use to get the simplified expression

Can someone help me just using basic mathematics did not help I tried it by myself but no luck I am pretty sure I have missed some assumptions

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    $\begingroup$ Can you reference the book? $\endgroup$
    – M. Farooq
    Mar 1 at 14:22
  • $\begingroup$ What are all those variables K'', K, K, Y', $N_{total}$ etc.? Also, please visit this page, this page and this one on how to format your posts competently with MathJax and Markdown. $\endgroup$
    – andselisk
    Mar 2 at 4:13
  • $\begingroup$ I can define all the constants and respective chemical equations then the question will be very long? $\endgroup$
    – Avon97
    Mar 2 at 4:57
  • $\begingroup$ Defined the constants now but $\ce{K''}$ is unknown $\endgroup$
    – Avon97
    Mar 2 at 6:03
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    $\begingroup$ You have not involved acidobasic conditions, affecting [Y^4-]/[EDTA total] and pH changes due titration. $\endgroup$
    – Poutnik
    Mar 2 at 7:25
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I finally found the answer to the question I have made a huge mistake in the testing equation

In order for a titration to happen the Formation constant $\ce{K}$ should be greater than or equal to $\ce{10^7}$

What I have written in my note book is

$\ce{K''_{MY^{n-4}}} = \ce{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}-α_{Y^{'}}}$

Which is incorrect it,

Should be like this

$\ce{K''_{MY^{n-4}}} = \ce{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}α_{Y^{'}}}$

The conditional formation constant of metal $\ce{M}$ should be multiplied by $\ce{α_{Y^{'}}}$ not subtracted

Substituting the equation for $\ce{α_{Y^{'}}}$ will give the simplified equation

$$\ce{α_{Y^{′}}= \frac{1}{(1+K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}])}}$$

$$\ce{K''_{MY^{n-4}} = \frac{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}}{(1+K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}])}} $$

But since

$$\ce{K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] >> 1 }$$

$$\ce{K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] + 1 \approx K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] }$$

Which gives the following final simpified experssion

$$\ce{K''_{MY^{n-4}} = \frac{K_{MY^{n-4}}α_{M^{n+}}}{K_{NY^{m-4}}α_{N^{m+}}[N_{total}]}} \geq 10^7 $$

Finally the testing equation is sloved for selective titrations

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