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:
The asymptote is parallel to the x-z plane
We pick the x-z plane as the plane for our projection
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:
Remove the denominators by multiplying each row by the
appropriate amount.
Make C2 constant by performing C2 + 2C3 → C2
Make two of the entries in C3 constant by performing C1 - C3
→ C3
Eliminate the 1 in the \(1-\sqrt{D}\) term using C3 + C2/2
→ C3
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.
Exchange the first and third columns to put the matrix in
standard nomographic form.
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.
