Exercises: pH, Logs and [H+]

For these execises you are advised to make use of your own electronic calculator, however, log10 and antilog tables or pop-up calculator are available by clicking appropriately.

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

A 1 mol dm_3 solution of a strong acid (ideal, monobasic) will have a [H+] of 1 mol dm_3.

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:

Calculate the pH of mol dm_3 HCl.
mol dm_3 of a strong acid will have a pH of mol dm_3.

I calculated: so allowing for a little variation, your response is:

Format B

Example:

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

Created and maintained by Robert Lancashire and Andrew Pearson,
University of the West Indies, Mona Campus, Jamaica.
Comments to authors at Robert Lancashire and/or Andrew Pearson.