Exercise 1 solution

pH = -log10[H+], so [H+] = 10^(-pH).

A fall in pH of 0.1 multiplies [H+] by 10^(0.1) = about 1.259.

That is an increase of about 26 per cent. (A fall of 0.1 from 8.2 to 8.1 means [H+] goes from 10^-8.2 to 10^-8.1, a ratio of 10^0.1.) The water is still on the alkaline side of neutral (pH 7).