Suppose you have prepared two batches of a particular solution,
e.g. hydrochloric acid. Each solution has a different
concentration. You want to determine what the combined
concentration will be if you mix together a particular amount of
each. You can find the answer to this type of problem—and
several others— with this nomogram. In this case:

First, decide how many liters of each solution you will
combine together. Locate those values on the red axes.

Draw a line connecting those values. The line will cross
the thick
vertical axis at a certain height—mark that
height. The height ranges from 0 to 1, and represents the
percentage (by volume) of solution A in the final solution.

Find the concentration of each solution on
the blue axes. Follow the
guideline eminating from your chosen value until you reach
the height you found in the previous problem. (If there's no
guideline for your chosen value, you may interpolate.)

The horizontal separation between the two guidelines at that
height is the concentration of the final solution. (On a
printed nomogram, you could use a ruler or count grid
squares to measure the horizontal separation.)

This nomogram has six variables: two volumes, three
concentrations, and percentage. It is a compound nomogram made
of two independent nomograms:

Given any two of the variables ⟨volume_A, volume_B,
percentage⟩, you can solve for the missing one.

Given any three of the variables ⟨concentration_A,
concentration_B, percentage, concentration_final⟩, you
can solve for the missing one.

More economically, you can use certain combinations of
four variables to solve for the other two: You can start
with any four of the concentration/volume variables (as we did
here), or you can start with one volume, two concentrations,
and a percentage. Either way, you can solve graphically for
the two variables you don't know.

Postscript: It's illuminating to see how qualitative relations
in this experiment are borne out graphically. For example, as
the amount of solution B dwindles to zero, what happens to the
fraction of A in the final solution? Relatedly, what happens to
the concentration of the final solution?
Second, notice how the concentration of the final solution must
lie between the concentrations of the component solutions;
geometrically, this is because the two guidelines constraint the
maximum width corresponding to the final concentration.
Finally, observe that this problem must be symmetric
with respect to the labels A and B because the nomogram itself is
symmetric.

Appendix: How I derived this nomogram

First, I wrote out the equation I wanted to plot. Here, if \(A\)
and \(B\) are the concentrations of the two solutions, and \(a\)
and \(b\) are the amounts to be combined, and \(C\) is the
concentration of the result, then the relevant equation is:
$$C = \frac{a\cdot A + b \cdot B}{a+b}.$$
I didn't know any nomographic form that could accomodate all five
of these variables, so I attempted to make a variable substitution
of some kind. I conveniently found \(t \equiv \frac{a}{a+b} \), the
percentage of the solution made up of the first component;
substituting for \(t\) in the above equation, I obtained
$$C = t\cdot A + (1-t)\cdot B$$
which gets rid of the two "amount by volume" variables and
expresses the concentration of the final solution as a weighted
average of the component solutions' concentrations.
I recognized this as an instance of a nomogram with oriented
transparencywhose general form is
$$P(x_1) = Q(x_2,x_3) + R(x_2,x_4).$$
I plotted the nomogram-with-transparency and obtained the diagram you see above except for the red (amount) axes.
Those axes came as a later addition when I lamented the fact that
the nomogram itself wasn't very useful because \(t\) is not
necessarily a natural quantity to come across in a mixing
problem. I looked for a way to re-integrate amount of solution
into the picture— although I wasn't convinced it was
possible.
I wanted to plot a nomogram for \(t = \frac{a}{a+b}\), the
defining equation for \(t\). Rearranging slightly, this becomes:
\(-a + at + bt = 0.\)
To do so, I had to find a 3×3 matrix in standard nomographic
form. This means I had to find expressions to make the following
equation hold for all values of \(a, b, t\):
$$
\det
\begin{bmatrix}
f_1(a) & f_2(a) & 1\\
g_1(b) & g_2(b) & 1\\
h_1(t) & h_2(t) & 1\\
\end{bmatrix}
= -a + at + bt.
$$
Furthermore, since this nomogram had to align with the \(t\) axis
already drawn, the last row was already determined, with \(h_1(t)
= 0\) and \(h_2(t)=t\).
It turned out, to my surprise, that there was at least one set of
functions \(f_i\) and \(g_i\) which solved the equation, namely:
$$
\det
\begin{bmatrix}
a & 0 & 1\\
-b & 1 & 1\\
0 & t & 1\\
\end{bmatrix}
= -a + at + bt.
$$
Thus, I was able to plot the five original variables plus a sixth
joining variable in a single nomogram (!).