How is the selective 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:
$$\ce{N^m+ + Y^4- &lt;=&gt;  NY^{m-4}}$$
$$\ce{M^n+ + Y^4- &lt;=&gt;  MY^{n-4}}$$
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 &gt; 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
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
$$\ce{K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] &gt;&gt; 1 }$$
$$\ce{K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] + 1 \approx K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] }$$
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