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When we find the pH of a solution, do we use normality or molarity? My teacher said that if the compound is present by itself, we can use normality but in all other cases we should use molarity. Can you explain why?

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Current definition implies that $\mathrm{pH}$ is a function of relative activity. Originally, the amount concentration of $\ce{H+}$ in $\pu{mol L-1}$ was proposed, which is also often used these days as an approximation [1].

$\mathrm{pH}$ was originally defined by Sørensen in 1909 … in terms of the concentration of hydrogen ions (in modern nomenclature) as $\mathrm{pH} = −\lg(c_\ce{H}/c^\circ)$ where $c_\ce{H}$ is the hydrogen ion concentration in $\pu{mol dm–3},$ and $c^\circ = \pu{1 mol dm–3}$ is the standard amount concentration. Subsequently …, it has been accepted that it is more satisfactory to define $\mathrm{pH}$ in terms of the relative activity of hydrogen ions in solution

$$\mathrm{pH} = -\lg a_\ce{H} = -\lg (m_\ce{H}γ_\ce{H}/m^\circ)\tag{1}$$

where $a_\ce{H}$ is the relative (molality basis) activity and $γ_\ce{H}$ is the molal activity coefficient of the hydrogen ion $\ce{H+}$ at the molality $m_\ce{H},$ and $m^\circ$ is the standard molality. The quantity $\mathrm{pH}$ is intended to be a measure of the activity of hydrogen ions in solution. However, since it is defined in terms of a quantity that cannot be measured by a thermodynamically valid method, eq. 1 can be only a notional definition of $\mathrm{pH}.$

So, if the numerical value for normality coincides with the one for molarity, and one can omit the activity in favor of molarity, i.e. when both equivalence factor $f_\mathrm{eq}$ and activity coefficient $γ$ are approx. 1, then yes, normality can also be used to estimate the $\mathrm{pH}.$

I'm not sure how to interpret "present by itself" part of your question, but if it means there is a single solute, then it seems to be incorrect. The simplest counter example would be a solution of di-, tri- and polybasic acid/base, say $\ce{H2SO4},$ for which the equivalent concentration is double the value of amount concentration.

References

  1. Buck, R. P.; Rondinini, S.; Covington, A. K.; Baucke, F. G. K.; Brett, C. M. A.; Camoes, M. F.; Milton, M. J. T.; Mussini, T.; Naumann, R.; Pratt, K. W.; et al. Measurement of pH. Definition, Standards, and Procedures (IUPAC Recommendations 2002). Pure and Applied Chemistry 2002, 74 (11), 2169–2200. DOI: 10.1351/pac200274112169. (Free Access)
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  • $\begingroup$ So in every scenario we would be using Normality rather than Molarity? If so then what do we use Molarity for? $\endgroup$ – Nate william Dec 13 '19 at 14:13
  • $\begingroup$ @Natewilliam No, you got it backwards, it's amount concentration, or molarity, that is used in the absence of activity. $\endgroup$ – andselisk Dec 13 '19 at 14:26
  • $\begingroup$ Isn't it the same? If there's no activity, then normality is equal to molarity ( as Normality = Molarity*activity(or n-factor) ) So isn't it correct to say that we always use normality for any sort of calculation rendering molarity unnecessary? $\endgroup$ – Nate william Dec 13 '19 at 14:39
  • $\begingroup$ @Natewilliam No. Normality is $c/f_\mathrm{eq}.$ If $f_\mathrm{eq} = 1$ (monobasic substance), then the numerical values of normality and molarity are the same. Activity has nothing to do with $f_\mathrm{eq}$ (or $n$-factor, s you call it), it's an unrelated quantity. $\endgroup$ – andselisk Dec 13 '19 at 14:52
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pH is not defined versus normality or molarity of an acid: Whatever the normality or the molarity, the pH is defined from the activity (or the concentration) of the ions H+. A given value of the normality or of the molarity does not give you the activity (or the concentration) of H+. So it does not allow you to calculate the pH.

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    $\begingroup$ The statements "pH is not defined versus normality or molarity" and "pH is defined from the activity (or the concentration)" seem to contradict each other as both normality or molarity are concentrations. You either have to be more specific, or stick with only one nominal definition. However, if you do the latter, this won't answer the question. Besides, I don't see how this adds anything to an existing answer. $\endgroup$ – andselisk Dec 13 '19 at 12:44
  • $\begingroup$ I think Maurice’s point is that OP seems to be referring to normality or molarity of an acid, not of protons. For a strong acid, the molar amount of protons is equal to normality of acid (which is all in the form of conj base). For a weaker acid, that is not the case, so pH cannot be calculated using only the acid molarity or normality. $\endgroup$ – Andrew Dec 13 '19 at 18:36

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