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When I learned about pH approximations in school, I was told that each approximation fits to a certain acidity strength/pKa range. Those ranges are grouped in common categories (pKa borders are only approximately, as they change from book to book):

  • very strong acids: $\mathrm{pK}_a < 0$
  • strong acids: $0 \le \mathrm{pK}_a < 4$
  • medium strong acids: $4 \le \mathrm{pK}_a < 8$
  • weak acids: $8 \le \mathrm{pK}_a < 14$
  • very weak acids: $\mathrm{pK}_a \ge 14$

From this point on, the arguments were made and approximations for certain strengths of acids were evaluated. But is it possible to choose the "right" approximation only based on acidity strength? Do those equations really only apply to it's categories?

This question originates in my wish to implement a program on my calculator that computes the pH for a given concentration and pKa value. I thought it might be wrong to simply ask for the pKa value and choose one of the approximations solely based on that.

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Preface

It is being assumed throughout this post that the concentration c equals the activity a. This is of course not true for concentrations above $\approx 10^{-3} \ldots 10^{-2}~\mathrm{mol\,L^{-1}}$ but it is common practice in schools and universities as long as the exercise is not very specifically about activities. $$\mathrm{pH} = -\lg a(\ce{H+}) \xrightarrow{a=c} \mathrm{pH} = -\lg [\ce{H+}]$$ The other general assumption is that temperature is $22~^\circ \mathrm{C}$ and therefore $k_\text{W} = 10^{-14}~\mathrm{mol^2\,L^{-2}}$.

Given error contours are not "real measurements vs. calculations" but absolute differences between approximation n and the cubic equation that is not more simplified than using concentrations instead of activities.


General

If you have a calculator that can solve cubic equations, you don't need to simplify. In addition you only would have to remember one equation, no matter which pKa or concentration occurs (see above for real world restrictions).

Solve the following equation for x, with $x=[\ce{H+}]$: $$x = \underbrace{\frac{k_\mathrm{W}}{x}}_{[\ce{OH-}]} + \underbrace{\frac{[\ce{HA}]_0 k_\mathrm{a}}{x+k_\mathrm{a}}}_{[\ce{A-}]}$$ or as a single equation (that is harder to remember): $$x^3+x^2 k_\mathrm{a}-x([\ce{HA}]_0 k_\mathrm{a}+k_\mathrm{W})-k_\mathrm{a} k_\mathrm{W} = 0$$

Results from this equation will be the reference pH values for the following comparison of pH approximations.

To calculate the pH for any monoprotic(?) base, simply replace [$\ce{H+}$] with [$\ce{OH-}$], $k_\mathrm{a}$ with $k_b$ and of course [$\ce{HA}$] with [$\ce{B}$].


Approximation 1 and 2

For very strong acids the approximation is that they are completely dissociated, which makes the resulting equation very easy: \begin{align} x &= [\ce{HA}]_0\\ \rightarrow \mathrm{pH} &= -\lg [\ce{HA}]_0 \end{align} When am I able to use this approximation with an error that is smaller than $\Delta \mathrm{pH} \leq 0.1$? It's the blue area in the picture below ($\mathrm{pK_c} = -\lg [\ce{HA}]_0)$.

enter image description here

The simple rule is: $6.7 \gtrsim \mathrm{pK_c} \gtrsim \mathrm{pk_a} + 0.5$.

Including the hydroxide ions from the autoprotolysis of water makes it a little bit more complicated and yields a quadratic equation in x: \begin{align} x &= [\ce{OH-}] + [\ce{HA}]_0 = \frac{k_\mathrm{W}}{x} + [\ce{HA}]_0\\ \rightarrow \mathrm{pH} &= -\lg \left(0.5 [\ce{HA}]_0 + \sqrt{0.25 [\ce{HA}]_0^2 + k_\mathrm{W}} \right) \end{align}

enter image description here

This formula holds within: $(\mathrm{pK_c} \gtrsim \mathrm{pk_a} + 0.5) \vee (\mathrm{pK_c} \gtrsim 7.4)$.

It's not surprising, that the range of usefulness increases over the whole are of diluted concentrations. And that the high pKa values are also covers is only due to the fact that the pH is 7 everywhere upon concentrations that are below $10^{-7}\ldots10^{-8}$.


Approximations 3 & 4

The following approximations are used for weak acids. It's the second easiest to remember, so it's used quite often. The assumption is, that $x \gg k_\mathrm{a}$. \begin{align} x &= \frac{[\ce{HA}]_0 k_\mathrm{a}}{x}\\ \rightarrow \mathrm{pH} &= -\lg \sqrt{[\ce{HA}]_0 k_\mathrm{a}} = 0.5 (\mathrm{pK_c} + \mathrm{pk_a}) \end{align}

enter image description here

It's application range is: $(\mathrm{pK_c} < 13.7 - \mathrm{pk_a}) \wedge (\mathrm{pK_c} < \mathrm{pk_a} - 0.6)$.

It is obvious, that the approximation is useful in a small area where the acid is not too diluted and it's pKa value is not too low. Seems to fit well for weak acids.

Again, what happens if $k_\mathrm{W}$ gets involved involved?

\begin{align} x &= \frac{k_\mathrm{W}}{x} + \frac{[\ce{HA}]_0 k_\mathrm{a}}{x}\\ \rightarrow \mathrm{pH} &= -\lg \sqrt{[\ce{HA}]_0 k_\mathrm{a} + k_\mathrm{W}} \end{align}

enter image description here

Again, the range expands over very weak acids and strong diluted solutions as expected: $(\mathrm{pK_c} < \mathrm{pk_a} - 0.6) \vee (\mathrm{pK_c} > 14.2 - \mathrm{pk_a})$.


Approximation 5

The last approximation is for strong acids that are not diluted. The amount of protons in solution is equivalent to the amount of dissociated acid but this time with taking $k_\mathrm{a}$ into account. As only the influence of the autoprotolysis of water is neglected, it is "the most accurate" approximation for high to medium concentrations. This can easily be seen as it includes the scope of both approximations, 1 and 2. It is again a quadratic equation in x.

\begin{align} x &= \frac{[\ce{HA}]_0 k_\mathrm{a}}{x+k_\mathrm{a}}\\ \mathrm{pH} &= -\lg(-0.5 k_\mathrm{a} + \sqrt{0.25 k_\mathrm{a}^2 + [\ce{HA}]_0 k_\mathrm{a}}) \end{align}

enter image description here

It can be used if: $(\mathrm{pK_c} < 6.7) \wedge (\mathrm{pK_c} < 13.7 - \mathrm{pk_a})$.


Now there are all those approximations, with all their respective acceptable ranges but it's quite hard to compare five diagrams with another. I made two further pictures that show 1) which approximation is the best per (pkc,pks)-pair and 2) which approximation is the easiest to remember per (pkc,pks)-pair.

1) The best -- which function yields the lowest error in comparison with the exact equation?

enter image description here

2) The easiest -- which function yields the lowest error in comparison with the exact equation and if there is more than function with $\Delta \mathrm{pH} < 0.1$, which one is the easiest to remember?

enter image description here


So ... the pka ranges kind of resemble when to use which function but in the end it's a little bit more complicated. If $k_\mathrm{W}$ is neglected, the low-error-area (LEA) is quite small, only approx. 5 is able to yield acceptable values for a big range of concentrations and pka values. Approximations 1 and 3 should be used with care, as their LEAs are quite fancy. For approximations 1 and 3 it's often better to switch to their low-concentration counterparts 2 and 4 to be on the safe side.

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  • $\begingroup$ -1 for thinking that you can calculate to an error of < 0.01 pH units up to a concentration of 1 M without considering activity. pH = -log (H+ activity), Ka = a(H3O+)a(A-)/a(HA)a(H2O), Kw=a(H3O+)a(OH-)/(a(H2O))^2 $\endgroup$ – DavePhD Mar 31 '16 at 19:06
  • $\begingroup$ @DavePhD Thank you for your comment. Is your problem about the very wide pKc scale or about the 0.01 error contour in the images? You might have noted, that this Q/A is not about the calculation of "real" pH values using acitivities but about which of the approximation yields the lowest errors for a certain (pkc,pka)-pair with respect to the "exact" equation. This is all within the very common treatment of schools (and also universities) to ignore the activity. I am well aware, that those assumptions are wrong for concentrated solutions and I have mentioned that a few times here on chem.se. $\endgroup$ – pH13 - Yet another Philipp Mar 31 '16 at 20:10
  • $\begingroup$ My criticism corresponds to the high concentration region of the graphs (0.01-1 M) where error would definitely exceed 0.01 units and even 0.1 units close to 1M. Also the statement "no matter which pKa or concentration occurs" makes it sound like the cubic equation would give an exactly correct answer at all concentrations. I don't seen how someone reading the question or answer would think they aren't about approximating real pH. $\endgroup$ – DavePhD Mar 31 '16 at 20:25
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    $\begingroup$ I don't see why it is out of the scope of the question. In a previous millennium, my research and publications involved determining the pKas of amino acid groups in proteins by NMR. The protein solutions would be concentrated in terms of mass fraction, but still much less than 1M, and I would consider the decreased concentration of water. Have I ever seen a pH homework exercise that considers water concentration? I'll try to find one. My college physical chemistry text (Levine) included water activity in the Ka and Kw definitions and then explained approximations from there. $\endgroup$ – DavePhD Apr 1 '16 at 15:06
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    $\begingroup$ I like the way the issue is discussed here: "As long as we are talking about diluted solutions we may safely assume water concentration is constant, and error introduced into our calculations will be rarely higher than the precision of dissociation constants used in calculations" chembuddy.com/?left=pH-calculation&right=water-ion-product $\endgroup$ – DavePhD Apr 1 '16 at 17:08

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