Logs: How to Derive Henderson-Hasselbalch and Calculate pH Without a Calculator
Calculating logs without a calculator is not as difficult as you might expect it to be. First, you should be familiar with the following log rules:
log (xy) = log x + log y
log (x/y) = log x - log y
log (x)^y = y(log x)
You can use these rules to derive the Henderson-Hasselbalch equation from the Ka expression for an acid:
HA -> A- + H+
Ka = [A-][H+]/[HA]
Taking the -log of both sides gives us:
-log Ka = -log {[A-][H+]/[HA]}
The left side is the definition of pKa:
pKa = -log {[A-][H+]/[HA]}
The right side can be re-written using the log rules:
pKa = -log [A-] - log [H+] - (-log [HA])
The second right side term is the definition of pH:
pKa = -log [A-] + pH + log [HA]
Solving for pH:
pH = pKa + log [A-] - log [HA]
Using the log rules in reverse gives us the equation as it is normally written:
pH = pKa + log [A-]/[HA]
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You should also know the common base 10 logs. For example:
log 1/100 = -2
log 1/10 = -1
log 1 = 0
log 10 = 1
log 100 = 2
and one more that is very helpful for pH estimations: -log 0.5 = 0.3
So let’s say that I’m working on a pH problem. I find out that my [H+] is equal to 3.6 x 10^-5. What is the pH? We don’t have a calculator on any of the pre-health tests, so we need to be clever about this. I don’t know what the -log of 3.6 x 10^-5 is. But here’s what I do know:
-log (1 x 10^-5) = 5
-log (10 x 10^-5 = 1 x 10^-4) = 4
So, the -log (3.6 x 10^-5) must be some number between 4 and 5. That is usually sufficient to solve the problem. Your answer choices might be 3.4, 4.4, 5.4, and 6.4, and you will of course pick 4.4.
If necessary, we can do even better than this. Say two of our answer choices are 4.2 and 4.5. How do we pick between them? (Note that this situation should not come up on your test, but just in case it does, you’ll know what to do!)
Here’s where knowing that -log 0.5 = 0.3 comes in useful. We know that -log (5 x 10^-5) must be 4.3, because we already memorized that -log (0.5) = -log (5 x 10^-1) = 0.3. Since we want the -log (3.6 x 10^-5), it must be a number that is close to 4.3, but a bit higher (because 3.6 is between 1 and 5, then the pH must be between 4.3 and 5). So we can eliminate 4.2, and 4.5 must be the correct answer. (The actual pH value is 4.44.)
