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.
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
[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
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