$\begin{aligned}A W &\rightarrow W+A\\H_2 O &\rightarrow WQ\end{aligned}$

The amount and the change: $\begin{aligned}A W,W,A,Q&=a,b,c,d\\&\rightarrow -x,+x,+x,0\\&\rightarrow 0,+y,0,+y\end{aligned}$

The equilibrium equations:

$\begin{aligned}K &= \frac{(b + x + y) (c + x)} {a - x} \\K_w &=(b + x + y) ( d + y ) \end{aligned}$

To Solve, $\begin{aligned}y= \frac{K_w (c + x)}{K (a - x)} - d\\K_w= (b - d +x+\frac{K_w (c + x)}{K (a - x)})\frac{K_w (c + x)}{K (a - x)}\\k \equiv \frac{K_w}{K}\\(b - d + x)(a - x) +k(c+x)=(b - d) a + k c+(a - b + d + k) x-x^2\\K_w=\left(\frac{(b - d)a + k c+(a - b + d + k)x-x^2}{a - x}\right)\frac{K_w(c + x)}{K(a - x)}\\K=\left(\frac{(b - d) a + k c+(a - b + d + k) x-x^2}{a - x}\right) \frac{c + x}{a - x}\end{aligned}$

… (1)

Concentration to be positive means x between zero of as well as between $0 = x^2 − (a - b + d + k) x-((b - d) a + k c-c<x<a$

Also the answer remains whenever $b - d$ remains

When $k$ is small, equation (1) turns into the typical rational for solving $x$

$K=\frac{(b-d + x)(c + x)}{a - x}$

Further when $d=\frac{a}{2}, b=c=0$ , this is the case for mid point tiltration, then equation turns into

$K= \frac{(-\frac{a}{2} + x) x}{a - x}$

where $W$ concentration is $-\frac{a}{2} + x$, and its solution is almost $\frac{a}{2} + K$ leading to $W$ concentration is $K$

When $d=a, b=c=0$, aka the equivalent point, equation (1) turns into:

$\begin{aligned}K=\left(\frac{-a^2 +(2 a + k) x-x^2}{a - x}\right) \frac{x}{a - x} = \left(\frac{k x - (a - x)^2}{a - x}\right)\frac{x}{a - x} = \frac{k x^2}{(a - x)^2} - x\\y= \frac{k x}{a - x} - a\end{aligned}$

… (2)

Note that at equivalent point, $k$ cannot be omit even it is small because $x$ cannot be negative in this case because the concentration of $A$ would be negative. Define $z=a-x$ then

$\begin{aligned}K=\frac{k (a - z)^2}{z^2} + z - a\\y=\frac{k a}{z} - k - a\end{aligned}$

… (3)

Because $z$ is near zero so 3rd power of $z$ can be omit, then

$\begin{aligned}K z^2=k(a-z)^2+z^3-a z^2 \approx k(a-z)^2-a z^2\\\sqrt{K + a}{k} z = a - z\\z = \frac{a}{1 + \sqrt{\frac{K + a}{k}}}\\y= k\left(1 + \sqrt{\frac{K + a}{k}}\right) - k - a =\sqrt{ k (K + a)} - a\\x = a - \frac{a}{1 + \sqrt{\frac{K + a}{k}}}\end{aligned}$

So the $W$ concentration is

$\begin{aligned}W = x + y = \sqrt{k(K + a)} -\frac{a}{1 + \sqrt{\frac{K + a|}{k}}}\end{aligned}$

When $K$ tends to infinite, $W$ will be $\sqrt{K_w}$ When $a$ tends to infinite, $W$ will be 0