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I'm a high school student, and I'm learning about acid dissociation equilibrium.

I wanted to make the idea of Le Chatelier's principle rigorous in math, so I decided to take to it with some calculus. It was all fine with just the acid equilibrium:

$$\begin{align} K_a = \frac{\ce{[H3O+][A-]}}{\ce{[HA]}} \end{align}$$

My thought process was to call $\ce{[H3O+]}$ $c_1$, to call $\ce{[A-]}$ $c_2$, and to let $c_1$ increase by some small $dc$. By Le Chatelier's principle the backwards reaction will be favoured, and both $c_1$ and $c_2$ will change by some other $dc_2$. In this light, the total change in $c_1$ would be $dc_1 = dc + dc_2$.

Plugging into the equation for $K_a$, noting that $K_a$ is constant throughout the changes in each concentration, and observing that $\ce{[HA]} = c_0 - c_2$ (where $c_0$ is the concentration of $\ce{HA}$ before any dissociation), you can get an expression for the change in $c_2$ in terms of the initial change in $\ce{[H3O+]}$, namely:

$$\begin{align} dc_2 = -\frac{c_2 ^2}{c_0 K_a + c_2 ^2}dc \end{align}$$

This can be used to find the final change in $c_1$:

$$\begin{align} dc_1 = -\frac{c_0K_a}{c_0 K_a + c_2 ^2}dc \end{align}$$

This is (pretty much) a system of differential equations. Now I'm not that experienced with systems of differential equations, but I knew enough to solve this particular system.

However, I then tried the same thing with the ionic product also taken into account:

$$\begin{align} K_w = \ce{[H3O+][OH-]}. \end{align}$$

Call $\ce{[OH-]}$ $c_3$, and change it by $dc$. Long story short, this makes $c_2$ change by some $dc_2$, $c_3$ change by some $dc_3 = dc + dc_3'$ (where $dc_3'$ is the amount the concentration changes strictly after $dc$ has been added) and it makes $c_1$ change by some $dc_1 = dc_2 + dc_3'$. Plugging all of this into the two equilibrium expressions as before gets a system of three differential equations:

\begin{align} \frac{dc_1}{dc} &= -\frac{ac_1/c_2^2}{(c_1+c_3)(1+a/c_2^2)-c_3}; \\ \frac{dc_2}{dc} &= \frac{c_1}{(c_1+c_3)(1+a/c_2^2)-c_3}; \\ \frac{dc_3}{dc} &= \frac{ac_3/c_2^2}{(c_1+c_3)(1+a/c_2^2)-c_3}, \end{align}

where $a = c_0K_a.$

Problem is, this system of differential equations looks... ahem viciously unsolvable. So much so that I doubt that the process I took to get to it was right. So is the way I'm doing this correctly, or is there a more efficient and less calculus-y route? Also, what is this field of study even called? I couldn't find any resources on these types of calculations, which is why I'm on here.

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    $\begingroup$ I think you overcomplicate it by differential equations. First step is calculation of equilibrium conditions of the weak acid + water system. The second step is adding some strong acid and calculation new equilibrium conditions, using dissociation constants of the acid and water, mass and charge balance in ordinary algebraic equations. The practical evaluation of acidobasic systems are often a trade off between accuracy and simplicity, but this is a simple enough system. $\endgroup$ – Poutnik Oct 17 at 15:22
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    $\begingroup$ I agree with Poutnik, you are overcomplicating this. Differential equations are usually required to solve chemical systems that evolve over time (e.g. reaction kinetics). Equilibria by definition describe systems that have stopped changing (macroscopically at least). And you can handle the latter using only ordinary algebraic equations, namely equilibrium constants, mass balances and charge balances. Then you can put as many components as you want, and it will always work out. In your case, initially you have a weak acid $HA$ in water; then you add a strong acid $HX$. Try it out. $\endgroup$ – user6376297 Oct 17 at 16:38
  • $\begingroup$ @Poutnik Ahhh, you're right! I'm too used to differential equations from physics, so I assumed the way to do it was to add a little bit of $\ce{OH-}$, see the result and sum all of those additions. Of course, you can add all of it at once. Are there any benefits you know of to viewing the process quasistatically, or was this all a bit of a waste of time? In any case, thank you for pointing this out. $\endgroup$ – Baylee V Oct 18 at 7:16

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