# A six-variable nomogram for mixing solutions

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 transparency whose 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 (!).