Tips for teaching maths skills to our future chemists
Logarithms can be particularly problematic for students, but mastery of them is essential especially in physical chemistry. Undergraduates find logs difficult partly because they don't accept that log quantities are real numbers1 and partly because they use calculators instead of log tables so they don't realise that logs can be used to simplify calculations.2 Consequently, they often memorise the properties of logs and exponents without understanding the material.3 Here, after defining logarithms we will look at a couple of chemistry examples that illustrate their properties and their relevance.
If three numbers represented as a, b and c obey the relationship:
a = bc (i)
then we can also say that:
c = logba (ii)
The function logb is the logarithm to base b. Although logarithms can be defined to any base, we will consider here only logarithms to base 10 because these are conceptually easier to understand. In fact, by convention, when we refer to logarithms to base 10 we often drop the subscript so that log x is read as log10x. In this case the expression (i) becomes:
a = 10c
In which case:
c = log a
In other words, the logarithm to base 10 of a power of 10 is simply the power itself.
Most will have encountered the term pH, but a chemist should be familiar with its formal definition, ie:
pH = -log10([H+]/mol dm-3)
The definition is written to emphasise an important point, ie one can only take the logarithm of a dimensionless quantity, which in most cases is one without units. This is achieved by dividing the concentration [H+] by the unit mol dm-3. So if:
[H+] = 0.1 mol dm-3
[H+]/mol dm-3 = 0.1 mol dm-3 = 0.1
pH = -log100.1
This could be evaluated using a calculator but if we write 0.1 as 10-1, the expression becomes:
pH = -log1010-1
As we saw earlier, the logarithm is simply the power and so:
pH = -(-1)
The two negatives cancel, giving:
pH = 1
Now consider what happens if this solution is diluted by a factor of 10. We have:
[H+] = 0.1 mol dm-3/10 = 0.01 mol dm-3
[H+]/mol dm-3 = 0.01 mol dm-3 = 0.01 = 10-2
Using the same arguments as before:
pH = -(-2) = 2
Thus for a dilution factor of 10 we see an increase in pH of 1. In general terms a change of 10 on a linear scale results in a unit change on the logarithmic scale.
The Beer-Lambert law
When light of intensity Io passes through a length l of a solution of concentration c, the intensity I of the light which emerges is given by the equation:
log (Io/I) = εcl
where ε is the absorption coefficient. (The quantity cl is also known as the absorbance, A, of the solution). If we compare this with equation (i) and (ii), it follows:
Io/I = 10εcl
Taking the reciprocal of each side of this equation then gives:
I/Io = 10-εcl
If we consider a solution of haemoglobin of concentration 3.0 × 10-4 mol dm-3 and a path length of 2.0 cm, for which ε = 532 dm3 mol-1 cm-1 at 430 nm, we can calculate that:
εcl = 532 dm3 mol-1 cm-1 × 3.0 × 10-4 mol dm-3 × 2.0 cm = 0.32
log (Io/I) = 0.32
I/Io = 10-0.32 = 0.48
Note that this could be the appropriate point at which to introduce the term antilogarithm, though this may be better left to a discussion of inverse functions.
- T. Berezovski, MSc thesis, Simon Fraser University, 2004.
- M. D. Taylor, Mathematics Teacher, 1991, 84, 541.
- M. Gamble, Mathematics Teacher, 2005, 99, 66.