Suppose we reacted enough strong base to fully neutralise a weak acid, clearly at this point (the equivalence point) the $pH$ would not be $7$. This is because the reaction between the strong base and weak acid forms a salt, and this salt can dissociate to form a strong base, which affects $pH$. Now, suppose we added some more of the strong base, the influx of $OH^{-}$ ions would contribute to a higher $pH$. If we calculate the $pH$ based purely off the extra hydroxide ions we have added, ignoring the weak base, would the error be negligible?

  • $\begingroup$ Could you provide some insight as to why the effect gets less with adding more of the strong base? What interaction is there between the strong base and the carboxylate ion (weak base)? $\endgroup$
    – nav
    Mar 23 at 12:36
  • $\begingroup$ Chem+Math MathJax formatting: Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles. $\endgroup$
    – Poutnik
    Mar 23 at 13:42

1 Answer 1


The key is the fact that strongly alkalic pH due strong base presence practically suppresses the salt hydrolysis. Unless the acid is very weak.

Imagine there is $\pu{1 L}$ of the solution, containing $\pu{0.01 mol}$ of a monoprotic weak acid with $\mathrm{p}K_\mathrm{a}=5.0$ with added $\pu{0.02 mol}$ of $\ce{NaOH}$.

The resulting $\mathrm{pH}=12.0$.

The salt $\ce{NaA}$ is dissociated to $\ce{Na+}$ and $\ce{A-}$. The latter than undergoes hydrolysis

$$\ce{A- + H2O <=> HA + OH-}$$

with $$\log{\frac{[\ce{A-}]}{[\ce{HA}]}}= \mathrm{pH} - \mathrm{p}K_\mathrm{a}$$

$$\frac{[\ce{A-}]}{[\ce{HA}]}=10^{ \mathrm{pH} - \mathrm{p}K_\mathrm{a}}$$

$$[\ce{OH-(hydr)}]=[\ce{HA}]=10^{( \mathrm{p}K_\mathrm{a} - \mathrm{pH} )}[\ce{A-}]$$

There would be about $\pu{e-9 mol L-1}\ \ce{OH-}$ resulting from the salt hydrolysis ( $10^{-2} \cdot 10^{5-12}=10^{-9}$) which is negligible.

The salt would have an effect in resulting pH, but for reasons of affecting ionic strength and ion activity coefficients. I assume this is out of scope of your level.

  • $\begingroup$ I can work out that there is $10^{-2}$ mols of $OH^{-}$ and $3.16 ⋅ 10^{-4}$ mols of $H^{+}$. Then we are left with $9.68 ⋅ 10^{-3}$ mols of $OH^{-}$ which gives $pH = 11.986$. However from this can we not say that the amount of mols of $H{+}$ is $10^{-12}$, when before we clearly had more $H{+}$ than $OH^{-}$, and hence we were left with $0$ $H^{+}$? $\endgroup$
    – nav
    Mar 23 at 13:03
  • $\begingroup$ I used an approximation which is only valid for when the acid is in water on its own... I see my mistake. Why on Earth does this give me the same answer as your's? If my method was completely off. How did you determine the amount of hydroxide ions resulting from the hydrolysis? I think all of the salt dissolves into carboxylate ions and hydrogen ions. How do either of these result in hydroxide ions? $\endgroup$
    – nav
    Mar 23 at 13:22
  • $\begingroup$ I see, what about the other part of my comment? You have provided a lot of help so far, thank you a lot $\endgroup$
    – nav
    Mar 23 at 13:29
  • $\begingroup$ Everything you wrote makes sense. One final question, the equilibrium I am used to seeing for the dissociation of a weak acid is not what you have written, but rather your equilibrium with a hydroxide ion "subtracted" from both sides. Are they the same equilibria? $\endgroup$
    – nav
    Mar 23 at 13:45
  • $\begingroup$ They are two mutually dependent equilibrii with pKa(of HA) + pKb (of A-) = 14. The stronger acid HA is, releasing H+, the weaker base A- is, releasing OH-. $\endgroup$
    – Poutnik
    Mar 23 at 13:53

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