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# Solution Focused

“This math quantifies a dilution;
Molarity of new solution,
M2, can be found.
Shift equation around:
M1 times V1; over V2.  Done!”

This Twitter poem, originally posted 4 November 2019, discusses a common equation taught in General Chemistry, taking significant advantage of chemical shorthand to fit into the limerick structure.  One focus of an introductory chemistry course involves solution stoichiometry: the arithmetic governing reactions that take place in aqueous solution (in water).

“This math quantifies a dilution…”
Quantifying (calculating) what happens when an aqueous solution is watered-down, or diluted, involves a key equation, the terms of which will be defined subsequently: M1V1 = M2V2.

“Molarity of new solution, /
M2, can be found.”
Chemists use “molarity” as a convenient unit of concentration: how much of a solute of interest, represented in moles, will be present in one liter of a solution

Using the equation above, we compare the molarity and volume of a stock solution– properties of a reagent we could take off the stockroom shelf, denoted here as “solution 1”– to the molarity and volume of a new solution, denoted as “solution 2.”  Specifically, we can find the molarity of the new solution, represented correctly as M2 and in the poem as M2.  (As ever, I lament my inability to have used subscripts with the original post.)

“Shift equation around: /
M1 times V1; over V2.  Done!”
This is a strained set of lines: algebraic explanations are not poetic.  However, this is how I’d teach the concept in class, manipulating the variables of molarity (M) and volume (V).

Starting with the equation of interest (M1V1 = M2V2) and rearranging to solve for M2, we end up with M2 = (M1V1)/V2.  To get there, we “shift the equation around.”  The product of the molarity and volume of the original solution is in the numerator (“M1 times V1”), while the volume of the new solution (“V2”) is now in the denominator.  That completes our calculation (“Done!”).  The double meaning of “solution” is interesting to consider here, as we find the solution to an algebraic calculation that itself involves the characteristics of an aqueous solution.