# How is the selective complexometric titration testing equation simplified?

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

• Can you reference the book? Mar 1, 2021 at 14:22
• 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. Mar 2, 2021 at 4:13
• I can define all the constants and respective chemical equations then the question will be very long? Mar 2, 2021 at 4:57
• Defined the constants now but $\ce{K''}$ is unknown Mar 2, 2021 at 6:03
• You have not involved acidobasic conditions, affecting [Y^4-]/[EDTA total] and pH changes due titration. Mar 2, 2021 at 7:25

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