Is it possible to calculate the dissociation degree (DD) from pH and Ka without directly been given the initial acid concentration?

I found the following formula:

$$\text{DD} = \frac{K_\mathrm{a}}{10^\mathrm{-pH}}$$

Is this true?

  • $\begingroup$ Briefly: yes, it's possible for any polyprotic acid, using pH and Ka only; no, the formula for DD seems to be incorrect. If you read German, the Wikipedia article covers it all: de.wikipedia.org/wiki/Dissoziationsgrad. If not, I can put some sort of an answer based on that in a couple of hours. $\endgroup$ – andselisk Jan 10 at 13:17
  • $\begingroup$ I can't read german, but thanks. I got this equation from ostwalds law and pH = 1.2pka - 1/2log ca. However, this is probably only valid when the acid is considered very weak? $\endgroup$ – delivosa Jan 10 at 13:50
  • $\begingroup$ No, it's only valid for any monoprotic acid and weak polyprotic acids. $\endgroup$ – andselisk Jan 10 at 13:54

Even though the degree of dissociation $α$ depends both on the nature of the dissolved electrolyte (e.g. acid) and the concentration, for the homogeneous medium it is possible to determine $α$ by $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}$ only, without any auxiliary information such as initial concentration since $\mathrm{pH}$ is a function of concentration.

I would suggest you not to remember certain formulas, since they have been derived only for particular systems and cannot be universally applied. Instead, use the law of mass action, especially when it comes to multi-component systems and multi-step dissociating electrolytes.

The following is heavily inspired by the German Wikipedia article for the degree of dissociation. Consider all $n$ steps of the dissociation of the $n$-protic acid $\ce{H_nA}$:

$$ \begin{align} \ce{H_nA + H2O &<=> H_{n-1}A- + H3O+}&\quad &K_{\mathrm{a}1} \tag{1}\\ \ce{H_{n-1}A- + H2O &<=> H_{n-2}A^2- + H3O+}&\quad &K_{\mathrm{a}2} \tag{2}\\ &\ldots &&\ldots\\ \ce{H_{n-(m-1)}A + H2O &<=> H_{n-m}A^m- + H3O+}&\quad &K_{\mathrm{a}m} \tag{$m$}\\ &\ldots &&\ldots\\ \ce{H2A^{(n-2)-} + H2O &<=> HA^{(n-1)-} + H3O+}&\quad &K_{\mathrm{a}(n-1)} \tag{$n-1$}\\ \ce{HA^{(n-1)-} + H2O &<=> A^n- + H3O+}&\quad &K_{\mathrm{a}n} \tag{$n$} \end{align} $$

where $n\in\mathbb{N}$, $m\in\mathbb{N}$, $m < n$; $K_{\mathrm{a}m}$$m$-th dissociation constant. Dissociation degree $α$ for any component of the composition $\ce{H_{n-m}A^m-}$ formed at the $m$-th step is by definition

$$α_m = \frac{[\ce{H_{n-m}A^m-}]}{c_0}$$

where $c_0$ – initial concentration:

$$ \begin{align} c_0 &= [\ce{H_nA}] + [\ce{H_{n-1}A-}] + \ldots + [\ce{H_{n-m}A^m-}] + \ldots + [\ce{HA^{(n-1)-}}] + [\ce{A^n-}]\\ &= \sum_{m=0}^{n}{[\ce{H_{n-m}A^m-}]} \end{align} $$

Each concentration in this set can be found recursively from the dissociation constant:

$$K_{\mathrm{a}m} = \frac{[\ce{H_{n-m}A^m-}][\ce{H3O+}]}{[\ce{H_{n-(m-1)}A^{(m-1)-}}]}$$


$$ \begin{align} [\ce{H_{n-m}A^m-}] &= \frac{K_{\mathrm{a}m}[\ce{H_{n-(m-1)}A^{(m-1)-}}]}{[\ce{H3O+}]}\\ &= \frac{K_{\mathrm{a}m}K_{\mathrm{a}(m-1)}[\ce{H_{n-(m-2)}A^{(m-2)-}}]}{[\ce{H3O+}]^2}\\ &\ldots\\ &= \frac{\prod_{i=0}^m{K_{\mathrm{a}i}}[\ce{H_{n-m}A^{m-}}]}{[\ce{H3O+}]^{n-m}} \end{align} $$

so that for $α_m$ after rearranging sums of the products the following universal equation of two arguments ($K_\mathrm{a}$ and $\mathrm{pH}$) emerges:

$$α_m = \frac{[\ce{H3O+}]^{n-m}}{[\ce{H3O+}]^n + K_{\mathrm{a}1}[\ce{H3O+}]^{n-1} + K_{\mathrm{a}1}K_{\mathrm{a}2}[\ce{H3O+}]^{n-2} + K_{\mathrm{a}1}K_{\mathrm{a}2}\cdot\ldots\cdot K_{\mathrm{a}n}}\cdot\prod_{i=0}^{m}{K_{\mathrm{a}i}}$$

For the monoprotic acid $\ce{HA}$ ($n = 1$, $m=0$) the formula becomes much simpler:

$$α_0 = \frac{K_{\mathrm{a}}}{K_{\mathrm{a}} + [\ce{H3O+}]} = \frac{K_{\mathrm{a}}}{K_{\mathrm{a}} + 10^{-\mathrm{pH}}},$$

and if the acid is very weak ($K_{\mathrm{a}} \ll 10^{-\mathrm{pH}}$, e.g. $K_{\mathrm{a}} \ll \text{ca.}~10^{-7}$), only then you can use your formula:

$$α_0 \approx \frac{K_{\mathrm{a}}}{10^{-\mathrm{pH}}}$$

  • $\begingroup$ Pragmatically I think that this also means that pH < 5 or pH > 9 to avoid complications due to autodissociation of water (assuming that you want 1% accuracy...) $\endgroup$ – MaxW Jan 10 at 21:49
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    $\begingroup$ Thank you so much! This is such a clear answer. $\endgroup$ – delivosa Jan 12 at 8:58

The formula seems fine, depending on how DD is defined.

The calculation is based on the Henderson-Hasselbalch equation (also known as the Buffer eq.):

pH = pKa + log ([Base]/[Acid]) => [B]/[A] =10^(pH-pKa)

The ratio [B]/[A] is independent of the total concentration. The ratio B/A can also be expressed as the fraction of the total concentration present as Protonized acid (Ya):

pH = pK + log ((1-Ya)/Ya), where Y goes from 1 (all acid protonized) to 0 (all acid deprotonized)

A Bjerrum plot is often used to illustrate Ya as a function of pH

  • $\begingroup$ And is this only valid if you say that the acid is very weak? So doesn't dissociate very much? $\endgroup$ – delivosa Jan 10 at 13:40
  • $\begingroup$ It is valid for all acids, but (very) strong acids will for all practical purposes be completely dissociated (Ya=1). But it is only valid for diluted solutions. When the concentration is too high the calculated result will deviate. $\endgroup$ – FrankS Jan 10 at 14:38
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    $\begingroup$ The equation isn't valid for dilute solutions of a weak acid where the autoionization of water would need to be considered. $\endgroup$ – MaxW Jan 10 at 21:52

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