Exercises: pH, Logs and [H⁺]

These examples have few formats, but the values are generated by a random number generator. Each time that you reload the page using your web browser reload button, fresh values will be loaded and corresponding solution expected. Enter your solution in the Answer box, remembering to use a - (minus) to signify 'bar', if appropriate. Simple errors will trigger a text box indicating the likely source of error, unforeseen errors may trigger unforeseen responses.

Format A

Example:
Calculate the pH of 0.005 mol dm^_3 HCl.

0.005 mol dm^_3 of a strong acid will have a [H⁺] of 0.005 mol dm^_3.
[H⁺] = 5 x 10^_3 mol dm^_3.
^_log₁₀ [H⁺] = ^_(log₁₀ 5 x 10^_3)
^_log₁₀ [H⁺] = ^_(-3.6990) = 2.3010
therefore pH = 2.3

Notes on ^_(-3.6990) = 2.301
Log₁₀ 5 x 10^_3 = (log₁₀ 5 + log₁₀ 10^_3)
Log₁₀ 5 x 10^_3 = (0.6990 + ^_3) = -3.6990
^_(Log₁₀ 5 x 10^_3) = ^_(0.6990 + ^_3) = ^_(-3.6990)
The nuisance of ^_logs is that the characteristic and mantissa must be treated separately:
the positive mantissa must be subtracted from the negative characteristic, so that:

^_(0.6990 + ^_3)
= ^_(^_3 + 0.6690)
= ^{_ _}3 ^_ 0.6990
= +3 ^_ 0.6990
= +2.3010

Now try out a calculation yourself:

Format B

Calculate the concentration of a solution of strong acid with a pH of 4.6.

^_(log₁₀ [H⁺] = 4.6 (= ^_log₁₀ [strong acid])

therefore log₁₀ [H⁺] = ^_4.6 = -5.4

therefore [H⁺] = antilog -5.4 = 2.512 x 10^_5 mol dm^_3

URL: http://wwwchem.uwimona.edu.jm/courses/pH/phexerc.html