I’ve recently begun tutoring high school chemistry. Unfortunately, I’m quite rusty at the subject and, truth be told, it has never been one of my favourites. So I’ve been re-learning a lot of chemistry in the last little while to keep up with my students’ demands. I would begrudge it, if not for the fact that I really should be more comfortable with basic chemistry if I plan to continue a career in any area related to environmental science. So learn, I will.
In one of my lessons, I encountered a chemical equation that gave me and my student a little trouble when we tried to balance it. It’s not particularly complicated, and we happened upon the correct molar ratios eventually, but it highlighted what I felt was a deficiency in the way they’d been taught to balance equations at the high school level.
Here’s the chemical equation (not yet balanced):
Na3PO4(aq) + Ba(OH)2(aq) —> NaOH(aq) + Ba3(PO4)2(s)
And here’s the guidance, as provided by the lesson:
The third step is to show that atoms on both sides of the arrow are conserved. Remember that you cannot alter the formula of the compounds but you can alter the number or coefficients of the compounds or elements. A balanced equation has the smallest whole-number ratio of coefficients.
Now, though this provides the theory necessary to balance the equation, it stops short of actually providing a methodology for doing so. Instead, the expectation seems to be that students should “play around” with the coefficients until atoms are conserved (that is, until the same number of Na, Ba, etc are on each side of the equation).
In a lot of cases, that’s fine. But it’s hardly a scientific method, and as equations quickly become more complicated, it’s simply not sufficient. Fortunately, I remembered hearing something about an algebraic method for solving chemical equations a long time ago (we won’t get into quite how long ago). It’s not complicated. It goes like this:
Step 1: Assign each molecule a letter
A Na3PO4(aq) + B Ba(OH)2(aq) —> C NaOH(aq) + D Ba3(PO4)2(s)
Step 2: Create equations for each atom
For example, to balance sodium (Na), one can see that for every molecule A on the left of the equation, there must be three molecule Cs, so A = 3C. (Polyatomic ions can be considered as “one atom” for this purpose.)
Na –> 3A = C
PO4 –> A = 2D
Ba –> B = 3D
OH –> 2B = C
Step 3: Solve the equations
Now, there is not a unique solution to these equations, as you can always multiply all the terms by something to arrive at a different overall quantity of reactants/products. However, the ratios will remain the same. Thus, you can start by arbitrarily selecting a coefficient for any one of the values. I’m going to select “A = 1″. Once I plug that in, I end up with the following values:
If A = 1 :
Na –> 3A = C … C = 3
PO4 –> A = 2D … D = 1/2
Ba –> B = 3D … B = 3 /2
OH –> 2B = C
Oh no! I’ve got fractions! No worries — as I mentioned above, the actual values don’t matter: just the ratios. Simply multiply all of the coefficients by something that will kill the fractions (ideally, the lowest common denominator). In this case, that’s 2. So…
A = 1 x 2 = 2
B = 3/2 x 2 = 3
C = 3 x 2 = 6
D = 1/2 x 2 = 1
Step 4: Plug the coefficients back into the equation
2 Na3PO4(aq) + 3 Ba(OH)2(aq) —> 6 NaOH(aq) + Ba3(PO4)2(s)
This level of algebra should be understood by any student in grade 11 chemistry, so I don’t know why this method isn’t the first thing taught to them.
So I don’t know about you, but I’m through guessing.