Another area of difficulty for students is taking Kw into consideration in calculations. Consider the following example looking at the pH rising by one unit for every 10-fold dilution of a strong acid:

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Now, consider dilute hydrochloric acid solution [HCl] = 10^-3 mol dm^-3 so [H⁺] = 10^-3 mol dm^-3 has a pH of 3; a solution of  [HCl] = 10^-4 mol dm^-3 so [H⁺] = 10^-4 mol dm^-3 has a pH of 4;  [HCl] = 10^-5 mol dm^-3 has a pH of 5 and so on. Then what is the pH of a solution with [HCl] = 10^-8 mol dm^-3?

Some students may jump to the answer of 8. But given time to reflect, they will realise that this is not a sensible answer for a dilute acidic solution to be pH 8 as this is alkaline! Thinking more carefully...

The ionisation of water becomes more significant in very dilute solutions.

Water contains 10^-7 mol dm^-3 H⁺ ions.

pH of 10^-8 mol dm^-3 HCl. When the water is included, the concentration of H⁺ ions in the solution is the sum of H⁺ ions from water, 1.0 x 10^–7, and acid, 1.0 x 10^–8 , giving about 1.1 × 10^-7 mol dm^-3, ie, pH 6.96.

When the concentration of a strong acid ≤ 10^-5 mol dm^-3 , the ionisation of water needs to be considered.

There is further support for maths manipulations of powers of ten and also walking through which buttons students should know on their calculators in the Maths skills course.

Being able to estimate most calculations with logs in your head by remembering just a couple of logs can also be a useful tool. This technique is explored on the next page.