Can anyone please tell me how to calculate the amount of 1 N NaOH required to raise the pH of water from 6 to 7, 8, 9, 10, 11 & 11.5. Thank you in advance for your kind response.
For a simple calculation you may use these equations:
[H+]= 10-pH [HO-] = 10pH-14
where [HO-] means molar/normal concentration of HO- in the final solution.
But you have to consider also, if the problem is practical, that:
- pH 6 is due to CO2 absorption and the initial sample contains an unknown quantity of H2CO3/HCO3- buffer.
- a basic pH obtained by NaOH addition is not stable because the sample continuously absorbs CO2.
Note that sodium molar balance would write: CNaOH = [Na+] + [NaOH], where CNaOH is the nominal molar concentration of NaOH in aq. sol. (formality, to be precise). Because NaOH is a strong base, it should be highly dissociated in aq. sol., so that [NaOH] [H3O+] > [OH-]/100, and we can no longer neglect [H3O+] compared to [OH-]. Then we shall have: CNaOH = [OH-] - [H3O+] = (Kw/[H3O+]) - [H3O+], or [H3O+]2 + CNaOH·[H3O+] = Kw, where Kw = 1.0·10-14 M2 is the autodissociation constant of water. Therefore, [H3O+]/M = - (½)(CNaOH/M) + √{{Kw + (¼)(CNaOH)2}/M2}, and:
pH = - log10{[H3O+]/M} = - log10{- (½)(CNaOH/M) + √{{Kw + (¼)(CNaOH)2}/M2}}
For CNaOH = 0 M; last equation yields pH = 7.
From last equation, we have 10-pH + (½)(CNaOH/M) = √{{Kw + (¼)(CNaOH)2}/M2}, or:
CNaOH = {{(Kw/M2) - 10- 2·pH}/10- pH} M
For instance, for pH = 7.1, we predict CNaOH = 4.65·10-8 M.
Also note that aq. sol. of NaOH are obviously not pH buffered ― their experimental pH can possibly deviate or progressively drift due to extraneous factors ― e.g. impurities, container corrosion, or carbonation by atmospheric CO2. Carbonated sol. tend to deviate to lower pH.
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