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I present to you a fractional composition question that my professor refuses to answer because he encourages me to use equations without understanding them. Any help you can provide would be very, very appreciated.

The fractional composition of $\ce{HA}$ for the disassociation of weak acid $\ce{HA}$ is presented as: $$\text{fraction of HA in the form HA} \equiv\alpha_{\ce{HA}} = \frac{[\ce{HA}]}{\ce{[A-] + [HA]}}$$

The acid being $\ce{HA}$ should have a 1:1 ratio of $\ce{H+}$ and $\ce{A-}$ because they disassociated from the same acid $\ce{HA}$ where there is one mole of each. At least, this is what makes sense to me. The equation above only shows $\ce{[A-]}$ in the denominator and I figure I could replace that with $\ce{[H+]}$ if I wanted to. But it turns out I can't because it doesn't work. And I don't know why.

For example,

If there's an acid $\ce{HA}$ with a pH of 3, the concentration of $\ce{H+}$ in solution is $10^{-3}$. However, this concentration of $\ce{H+}$ does not mean that $\ce{[A-]} = 10^{-3}$. Why?

Here is a more in-depth example from my textbook:

enter image description here

Thank you for your time.

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    $\begingroup$ You can afford to replace [A-] by [H+] only in case if all H+ can be considered as becoming from acid dissociation, i.e. auto-disociation of water can be neglected. It is aceptable only if [H+] >> [OH-]. This condition is not met for just weakly acidic solutions with $\mathrm{pH} \approx 6+$, if the acid is too weak and/or its concentration is too low. The condition is not met either if other solution components affect ratio of H+ and A- concentrations. $\endgroup$
    – Poutnik
    Mar 17, 2022 at 9:20
  • $\begingroup$ Please don't use screenshots of text. Don't forget to properly cite your sources. The last screenshot looks like it's from Harris' Exploring Chemical Analysis. "My textbook" is meaningless. $\endgroup$
    – andselisk ♦
    Mar 17, 2022 at 13:18
  • $\begingroup$ @Poutnik You're right. I completely overlooked this. Thank you. $\endgroup$
    – big_armpit
    Mar 19, 2022 at 2:34

1 Answer 1

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In solution you have $\ce{HA}$, $\ce{A-}$, $\ce{H+}$ from dissociation of the acid and $\ce{H+}$ already in the solution. The last one conditions the dissociation of the weak acid.

\begin{array}{cccc} \ce{HA &<=> &H+ &+ &A-}\\ x && 10^{-5.31}-x && -x \end{array}

The equilibrium constant is

$$K_\mathrm{a} = \frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}.$$

You can see now, that an already presence of one of the species involved in this equation, tend to alter the concentration of the other since the $K_\mathrm{a}$ has to remain the same in every condition if $T$ remains constant.

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    $\begingroup$ Note that \ce{…} macro from mhchem is used strictly for chemical expressions, not math operators or variables. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Familiarize yourself with Which symbols are written in roman (upright) font and which are italicized? $\endgroup$
    – andselisk ♦
    Mar 17, 2022 at 12:40
  • $\begingroup$ I had a similar thought to this but HA being a weak acid does not completely disassociate and therefore will not have the same concentration as A-. Thus, they cannot both be denoted with X. Another issue is that in the example, [A-] is bigger than [H+]. Using your diagram, plugging x into 10^-5.31 - x would yield a negative concentration. I tried using the Henderson-Hasselbach equation to see if a base with the same A- as the acid was added to the solution. But it did not appear so. The other possibility is that a salt with the same A- was added. ..... $\endgroup$
    – big_armpit
    Mar 19, 2022 at 2:38
  • $\begingroup$ ... (continued) ... However, the math done to calculate the final [A-] uses only information about the acid. $\endgroup$
    – big_armpit
    Mar 19, 2022 at 2:42
  • $\begingroup$ @big_armpit try to calculate the "normal" dissociation and then set the $\ce{[H+]}=10^{-5.31}$ e calculate again the $\ce{[A-]}$ $\endgroup$
    – Andrea Pellegrini
    Mar 19, 2022 at 13:19

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