Three reactions and the change:

$\begin{aligned}HA \rightarrow H^+ + A^-\\-x,+x,+x\\BOH^- \rightarrow OH^- + B^+\\-y,+y,+y\\H_2O \rightarrow H^+ +OH^-\\-z,+z,+z\end{aligned}$

Equation for general initial concentration

Initial amount:

$HA,H^+,A^-,BOH,OH^-,B^+ = a_1,b_1,c_1,a_2,b_2,c_2$

The equilibrium equations: $\begin{aligned}K_1=\frac{(b_1 + z + x)(c_1 + x)}{a_1 - x}\\K_2=\frac{(b_2 + z + y)(c_2 + y)}{a_2 - y}\\K_w=(b_1 + z + x)(b_2 + z + y)\end{aligned}$

After rescale to dimensionless by $\sqrt{K_w}$ where all concentrations are divided by $\sqrt{K_w}$

Therefore, by page 化學反應正平方根(%E5%8C%96%E5%AD%B8%E5%8F%8D%E6%87%89%E6%AD%A3%E5%B9%B3%E6%96%B9%E6%A0%B9)

$\begin{aligned}k_1=\frac{(b_1 + z + x)(c_1 + x)}{a_1 − x}\\k_2= \frac{(b_2 + z + y)(c_2 + y)}{a_2 - y}\\1=(b_1 + z+ x)(b_2 + z + y)\end{aligned\begin{aligned}x= \frac{- (b_1 + z + c_1 + k_1) + \sqrt{(b_1 + z + c_1 + k_1)^2 - 4 ( b_1 + z ) c_1 + 4 k_1 a_1}}{2}\\y= \frac{-(b_2 + z + c_2 + k_2) + \sqrt{(b_2 + z + c_2 + k_2)^2 - 4 ( b_2 + z ) c_2 + 4 k_2 a_2}}{2}\\4=\left(z + b_1 - c_1 - k_1 + \sqrt{(b_1 + z + c_1 + k_1)^2 - 4 ( b_1 + z ) c_1 + 4 k_1 a_1}\right)\left(z + b_2 - c_2 - k_2 + \sqrt{(b_2 + z + c_2 + k_2)^2 - 4 ( b_2 + z ) c_2 + 4 k_2 a_2}\right)\end{aligned}$

… (2)

Special case of initial

$HA,H^+,A^-,BOH,OH^-,B^+= a_1,0,0,a_2,0,0$

Knowing that the reacting amount of $HA$, $x$, and of $BOH$, $y$ are almost the same, the balance equation of $HA + BOH \rightarrow H_2O + A^- + B^+$ can be applied.

$\begin{aligned}k_1 k_2= \frac{x^2}{(a_1 - x)(a_2 - x)}\end{aligned}$

So, $0= (\frac{1}{k_1 k_2} - 1) x^2 +(a_1 + a_2)x - a_1 a_2$

And $\begin{aligned}x = \frac{- (a_1 + a_2) + \sqrt{(a_1 + a_2)^2 + 4 \left(\frac{1}{k_1 k_2} - 1\right) a_1 a_2}}{2\left(\frac{1}{k_1 k_2} - 1\right)}\\H^+ = \frac{k_1(a_1 - x)}{x}\\ OH^- = \frac{k_2(a_2 - x)}{x}\end{aligned}$

When $a_1 = a_2 = a$, we have $\sqrt{k_1 k_2} = \frac{x}{a - x}$ so $H^+= \sqrt{\frac{k_1}{k_2}}$

The real $H^+$ is $H^+ = \sqrt{\frac{K_w K_1}{K_2}}$ and therefore $pH = 7 + \frac{1}{2} \log{\frac{K_2}{K_1}}$

Say, $a_1 < a_2$, this initial is equivalent to another initial $HA,H^+,A^-,BOH,OH^-,B^+ = 0,0,a_1,a_2- a_1,0, a_1$

so by (2) its equation is

$\begin{aligned}4=\left(z - a_1 - k_1 + \sqrt{(z + a_1 + k_1)^2 - 4 a_1 z}\right)\left(z - a_1 - k_2 + \sqrt{(z + a_1 + k_2)^2 - 4 a_1 z + 4 k_2 ( a_2 - a_1 )}\right)\end{aligned}$

When at equivalence point it is

$\begin{aligned}4=\left(z - a - k_1 + \sqrt{(z + a + k_1)^2 - 4 a z}\right)\left(z - a - k_2 + \sqrt{(z + a + k_2)^2 - 4 a z}\right)\end{aligned}$

and further $k_2$ tends to infinite, this equation becomes

$\begin{aligned}2=\left(z - a - k_1 + \sqrt{(z + a + k_1)^2 - 4 a}z\right)z\end{aligned}$

Range of z

To decide range of $z$,

$\begin{aligned}z + b_1 - c_1 - k_1 + \sqrt{(b_1 + z + c_1 + k_1)^2 - 4(b_1 + z)c_1 + 4 k_1 a_1} > 0\end{aligned}$

Case 1. $z + b_1 - c_1 - k_1 > 0$ then this is obviously true, meaning $z > c_1 + k_1 - b_1$

Case 2. $z + b_1 - c_1 - k_1 \leq 0$ then

$\begin{aligned}\sqrt{(b_1 + z + c_1 + k_1)^2 - 4(b_1 + z)c_1 + 4 k_1 a_1}> c_1 + k_1 - b_1 - z \geq 0\end{aligned}$

Therefore,

So it is $\begin{aligned}(b_1 + z + c_1 + k_1)^2 - 4(b_1 + z)c_1 + 4 k_1 a_1 > (c_1 + k_1 - b_1 - z)^2\\(b_1 + z + c_1 + k_1)^2 − (c_1 + k_1 - b_1 - z)^2 > 4(b_1 + z)c_1 - 4 k_1 a_1\\4(c_1 + k_1)(b_1 + z)> 4(b_1 + z)c_1 − 4 k_1 a_1\\4k_1(b_1 + z)> -4 k_1 a_1\\z > -a_1 - b_1\end{alignz > c_1 + k_1 - b_1$ or together to be $c_1 + k_1 - b_1 ≥ z > - a_1 - b_z > - a_1 - b_1$

Similarly $z > - a_2 - b_2$ so overall it is $z > - \min(a_1 + b_1 , a_2 + b_2 )$ . This is intuitive by thinking the initial being $a_1$ and $a_2$ move to the right side of the reactions totally.

Equation (2) by x

$\begin{aligned}z= \frac{k_1(a_1 - x)}{c_1 + x} - x - b_1 = - x - b_1 - k_1 + \frac{k_1(a_1 + c_1)}{c_1 + x}\\k_1 k_2 \frac{a_2 - y}{c_2 + y}= \frac{c_1 + x}{a_1 - x}\\\frac{a_2 - y}{c_2 + y}=\frac{c_1 + x}{k_1 k_2 a_1 - k_1 k_2 x}\\\frac{a_2 - y}{a_2 + c_2}=\frac{c_1 + x}{k_1 k_2 a_1 + c_1 + ( 1 - k_1 k_2 ) x}\\y= a_2 - \frac{(a_2 + c_2)c_1 +(a_2 + c_2)x}{k_1 k_2 a_1 + c_1 + (1 - k_1 k_2)x}= \frac{k_1 k_2 a_1 a_2 - c_1 c_2 - (k_1 k_2 a_2 + c_2) x}{k_1 k_2 a_1 + c_1 +(1 - k_1 k_2)x}\\1 = \frac{k_1(a_1 - x)}{c_1 + x}\left(b_2 - b_1 - x + \frac{k_1 a_1 - k_1 x}{c_1 + x} + \frac{k_1 k_2 a_1 a_2 - c_1 c_2 -(k_1 k_2 a_2 + c_2)x}{k_1 k_2 a_1 + c_1 +(1 - k_1 k_2) x}\right)\end{aligned}$

… (3)

Equation at equivalence point

At equivalence point, initial with $a_1,b_1,c_1,a_2,b_2,c_2=0,0,a,0,0,a$ this becomes

$\begin{aligned}1 = \frac{-k_1 x}{a + x}\left( -x - \frac{k_1 x}{a + x} + \frac{- a^2 - a x}{a + (1 - k_1 k_2) x}\right)\end{aligned}$

For case of large $k_2$ and $\frac{a}{k_1}$, the term $- \frac{k_1 x}{a + x} + \frac{- a^2 - a x}{a + (1 - k_1 k_2) x}$ is small compared with $-x$ so this equation becomes

$\begin{aligned}1 = \frac{- k_1 x}{a + x}(-x) = \frac{k_1 x^2}{a + x}\end{aligned}$

which is the conventional method in the textbook to find the concentration of $OH^-$.

For case of equal $k$ strength

$\begin{aligned}- x + \frac{- a^2 - a x}{a +(1 - k_1 k_2) x} = \frac{k_1 k_2 x^2 - ( a + x )^2}{a + (1 - k_1 k_2) x} = -\frac{(x + a)^2 - k^2 x^2}{a + (1 - k^2) x} = \frac{\left(\frac{-k x}{a + x} - 1\right)\left(\frac{-k x}{a + x} + 1\right)}{a + (1 - k^2)x}(x + a)^2\end{aligned}$

Therefore

$\begin{aligned}0= \left(\frac{- k x}{a + x}\right)^2 - 1 + \frac{- k x}{a + x} . \frac{\left(\frac{- k x}{a + x} - 1\right)\left(\frac{- k x}{a + x} + 1\right)}{a + (1 - k^2)x}(x + a)^2 \\= \left(\frac{- k x}{a + x} - 1\right)\left(\frac{- k x}{a + x} + 1\right)\left(1 + \frac{- k x}{a + x} . \frac{(x + a)^2}{a +(1 - k^2) x}\right)\\= \left(\frac{- k x}{a + x} - 1\right)\left(\frac{- k x}{a + x} + 1\right)\left(1 - k x \frac{a + x}{a +(1 - k^2)x}\right)\\= \left(\frac{- k x}{a + x} - 1\right)\left(\frac{- k x}{a + x} + 1\right)\left(\frac{x^2 + (a + k - \frac{1}{k})x - \frac{a}{k}}{(k - \frac{1}{k}) x - \frac{a}{k}}\right)\end{aligned}$

Knowing that concentration of $H^+$ being $\frac{- k x}{a + x}$ must be positive, it is 1 and $x = \frac{- a}{1 + k}$

The zeros of $x^2 +(a + k -\frac{1}{k})x - \frac{a}{k}$ are either positive or less than $- a$ meaning the concentration of $A^-$ is negative. Because otherwise:

$\begin{aligned}x =\frac{- \left(a + k -\frac{1}{k}\right) - \sqrt{a + k -\frac{1}{k})^2 + 4 a k}}{2} > - a\end{aligned}$

Then

$\begin{aligned}(a - k + \frac{1}{k})^2 - (a + k -\frac{1}{k})^2 > \frac{4 a}{k}\end{aligned}$

Then

$\begin{aligned}- 4 a\left(k - \frac{1}{k}\right)> \frac{4 a}{k}\end{aligned}$

Then $k < 0$, a contradiction