How is the selective titration testing equation simplified?
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