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If this is the outcome of a (too) simplified calculation formula for pH of hydroxide solutions, then here is the 'missing' part to correct this 'false' answer (or outcome):

pH = 14.0 - pOH = 14.0 - [-log(OH-)] (this is generally valid at room temperature 25oC)

  • simplified: pH = 14.0 - [-log(OH-)] = 14.0 + log(NaOH) = 14-[log(10-8)]= 6
  • almost (but sufficiently) correct: .... = 14.0 + log[(NaOH) + (1.0*10-7)]

    so by adding the amount of OH- ions, already present in pure water. This can NOT longer be ignored when NaOH concentration is very diluted, below 10-6 mol/L.

So when pHoutcome < 8 for dilute bases it should be 'recalculated' by addition of 10-7 before taking any log !

(The same is analogously valid for dilute acids when pHoutcome > 6)

pHoutcome = 6 for this NaOH concentration (which appears to have been (NaOH) = 1.0*10-8) is lower than 8 so is to be corrected as follows:

pH = 14.0 + log[(NaOH) + (1.0*10-7)] = 14 + log[1.0*10-8 + 1.0*10-7] =

= 14 + log[1.1*10-7] = 14 - 6.96 = 7.04 = 7.0

(and not 6 as in the simplified formula outcome)

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