# Dissociation of weak acids and bases

When you add a weak acid or base to water, it will dissociate very slightly. The extent of dissociation is characterized by the equilibrium constant of the dissociation reaction, which depends on the species of acid/base, the temperature, and so on.

This nomogram encodes the dissociation behavior. The three variables are the concentration of the substance before it dissociates (vertical axis), the concentration remaining after it dissociates (curved axis), and the equilibrium constant (horizontal axis). If you know any two of the variables, you can draw a straight line through them to find the third.

Note how the qualitative behavior of the nomogram reflects physical reality: since only the acid/base is present initially, its concentration can only decrease toward equilibrium (by dissociating) — never increase. This is encoded in the fact that equal initial and equilibrium concentrations are located at the same height in the image; whenever equilibrium concentration meets or exceeds the initial concentration (a physical impossibility), the line joining them has positive slope and hence never intersects the axis for the equilibrium constant.

Similarly, you can map the behavior as you hold the initial concentration fixed and follow the equilibrium constant K towards infinity (the middle of the image). You'll see that the equilibrium concentration diminishes to zero because large K values correspond to extremely favorable dissociation. Conversely, as K diminishes toward zero (far off on the right), the equilibrium concentration grows until it almost equals the initial concentration; low K values correspond to almost no dissociation.

### Creating the dissociation nomogram

The following reaction describes the dissociation of a monoprotic weak acid: $$HA_{(aq)} + H_2O_{(\ell)} \leftrightarrow H_3O^+_{(aq)} + A^-_{(aq)}.$$ The definition of the equilibrium constant for this reaction is $$K \equiv \frac{[H_3O^+][A^-]}{[HA]}$$, where square brackets denote concentrations at equilibrium. If a solution of pure undissociated acid is added to water, then at equilibrium the products will exist in equal concentrations, $X$, and the initial concentration $C$ of the acid will diminish by the same amount. Hence we have $$K = \frac{X^2}{C-X}.$$ You can create a nomogram for this equation — e.g. using $\begin{bmatrix}K & 0 & 1 \\ -X & 1/X & 1 \\ 0 & 1/C & 1\end{bmatrix}$ — however, the quantity $X$ is not as interesting to me as the quantity $C-X$, the concentration of the acid at equilibrium. Hence I'll change variables, letting $D\equiv C - X$ be the concentration at equilibrium. $$K = \frac{(C-D)^2}{D}$$ Taking square roots and rearranging, we get $$C -\sqrt{K}\sqrt{D} - D = 0.$$ This equation has the form of a nomogram with two parallel scales and one curved scale, which in general looks like $$F_1(u) + F_2(v)F_3(w) + F_4(w) = 0$$ and can be plotted using $$\begin{bmatrix}F_1 & 0 & 1\\ F_2 & 1 & 1 \\ \frac{-F_4}{F_3 + 1} & \frac{F_3}{F_3 + 1} & 1\end{bmatrix}.$$ In our particular case, we have $$\begin{bmatrix}C & 0 & 1\\ \sqrt{K} & 1 & 1 \\ \frac{D}{1 - \sqrt{D}} & \frac{-\sqrt{D}}{1 - \sqrt{D}} & 1\end{bmatrix}.$$ Unfortunately, there's a singularity at $D=1$. This is undesirable because there isn't anything special about the concentration $D=1$ in chemistry; our algebra is just pathological. We can get rid of this asymptote by performing a projective transform.

The parametric curve traced out by $D$ has an oblique asymptote along the line $\langle D, D\rangle$. For simplicity, I'll start by rotating the nomogram counterclockwise by 45 degrees so that the asymptote is vertical. The transformation matrix for a 45 degree rotation is $\frac{1}{\sqrt{2}}\begin{bmatrix}1 & -1 \\ 1 & 1\end{bmatrix}$ — but I'll ignore the scale factor since it doesn't affect the content of the nomogram. More explicitly, this rotation (ignoring the normalizing scale factor) alters the first two columns of the nomogram matrix in the following way $$x^\prime = x - y\qquad\qquad y^\prime = x + y,$$ so that we get $$\begin{bmatrix} C & C & 1 \\ \sqrt{K}-1 & \sqrt{K}+1 & 1\\ \frac{D-\sqrt{D}}{1-\sqrt{D}} & \frac{D+\sqrt{D}}{1-\sqrt{D}} & 1\\ \end{bmatrix}$$ Rotating was not strictly necessary, but it will make the perspective transform a little bit nicer.

### Using a perspective transform to eliminate asymptotes

A perspective transform is defined by a plane and a three dimensional (“vanishing”) point. It transforms two dimensional shapes embedded into the x-y plane by projecting them onto that plane. A perspective transform will eliminate an asymptote if, for example, the following three conditions hold:
1. The asymptote is parallel to the x-z plane
2. We pick the x-z plane as the plane for our projection
3. We pick a vanishing point that lies directly above or below the asymptote in the z-direction.
Accordingly, we pick a point like $\langle 1, -1, 1\rangle$ for our vanishing point and the x-z plane as our projection plane. Although I don't derive it here, the projective transform for that point and plane is $$x^\prime = \frac{x}{x-1}\qquad\qquad y^\prime = \frac{-x - y}{x -1},$$ which when applied to our nomogram yields: $$\begin{bmatrix} \frac{C}{C-1} & \frac{2C}{C-1} & 1\\ \frac{1+\sqrt{K}}{\sqrt{K}} & \frac{2\sqrt{K}}{\sqrt{K}}& 1\\ \frac{D-\sqrt{D}}{D-1} & \frac{-2D}{D-1} & 1. \end{bmatrix}$$ You can confirm, by taking the determinant and multiplying by the appropriate value to eliminate denominators, that the perspective transform preserved our nomogram's structure; it still represents the equation $C - \sqrt{K}\sqrt{D} - D = 0.$ Of course, this new form should be tidied up somewhat — in particular, at first glance the singularity at $D=1$ hasn't gone away.

To tidy up, we'll perform some elementary matrix operations:

1. Remove the denominators by multiplying each row by the appropriate amount.
2. Make C2 constant by performing C2 + 2C3 → C2
3. Make two of the entries in C3 constant by performing C1 - C3 → C3
4. Eliminate the 1 in the $1-\sqrt{D}$ term using C3 + C2/2 → C3
5. Now, alas, there are zeros in all columns but the first. Hence to put the matrix in standard nomographic form, we must normalize that column. Divide each row by the appropriate amount.
6. Exchange the first and third columns to put the matrix in standard nomographic form.
$$\begin{bmatrix} 0 & \frac{1}{C} & 1\\ \frac{1}{\sqrt{K}} & 0& 1\\ \frac{-1}{\sqrt{D}} & \frac{1}{D} & 1. \end{bmatrix}$$

Postscript: As you can see from the matrix form, the curves of our nomogram are a vertical line, a horizontal line, and a parabola. Also, we could have derived this matrix without using a perspective transform simply by normalizing the first column of our original matrix, then performing C3 - C2 → C3. However, I felt like the perspective transform was a useful enough trick for a tutorial in any case.

Addendum (11 May 2015): When dealing with especially weak substances, it is sometimes more convenient to simplify the original equation $K = \frac{(C-D)^2}{D}$ by assuming that the quantity $X \equiv C-D$ is very small relative to $C$ and $D$ so that the denominator becomes $K \approx \frac{(C-D)^2}{C}.$ (By making this simplification, you guarantee that you can solve for $X$ simply by taking a square root, rather than using, say, the quadratic formula.) Now, both equations have similar nomographic forms, namely $C -\sqrt{K}\sqrt{D} - D = 0$ and $C -\sqrt{K}\sqrt{C} - D = 0.$ Surprisingly, both the original and simplified equation correspond to nearly the same nomogram except the labels for the $C$ and $D$ curves are interchanged, and the $K$ axis changes sign.