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Consider a buffer solution containing a weak acid and its conjugate base ($c_\mathrm{total} = \pu{0.1 mol L-1}$): $$\ce{HA <=> H^+ + A^-}$$

When $\ce{H+}$ ions are added (from a strong acid), the position of the equilibrium shifts to the right and $\ce{A^-}$ combines with most of the $\ce{H+}$, preventing a significant decrease in $\mathrm{pH}$.

Along the same line, my text book says upon the addition of a strong base, the $\ce{OH-}$ react with $\ce{H+}$ in the reaction mixture and reduce their concentration, causing the $\ce{HA}$ to ionize and replace most of them, which prevents a significant decrease in $\mathrm{pH}$.

However, If you add $\ce{OH-}$ ions from a strong base, say $c = \pu{0.01 mol L-1}$ to a buffer with $\mathrm{pH} = 5$, there obviously aren't enough $\ce{H+}$ ions in the reaction mixture to react with the base. How will the buffer function in this case?

I can think of two ways:

  1. The $\ce{OH-}$ ions will keep depleting $\ce{H+}$ from the reaction mixture, driving the equilibrium to the right until all $\ce{OH-}$ ions are neutralized, and then the system will try to regain equilibrium.
  2. The $\ce{OH-}$ ions react with $\ce{HA}$ instead, and then the partial increase in pH occurs from the shifting of the equilibrium to the left.

Is one of these correct? Or would the buffer just not work at all if $\ce{OH-}$ added is more than $\pu{E-5 mol L-1}$?

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    $\begingroup$ The principle is simple. All equilibrium constants are to be honoured simultaneously, if the solution is in equilibrium. It includes all respective acidity/basicity constants and water auto-dissociation constant. // You confuse equilibrium and reaction quotient. A reaction is at equilibrium when reaction quotient is equal to the equilibrium constant. $\endgroup$
    – Poutnik
    Commented Jul 6, 2021 at 13:38
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    $\begingroup$ You may find the answers in here useful. $\endgroup$
    – Mathew Mahindaratne
    Commented Jul 6, 2021 at 21:24
  • $\begingroup$ Many confrontation-like questions either A either B lead to compromise-like answers Both A AND B. OH- reacts with both H+ AND HA, simultaneously honouring all equilibrii expressed by Ka, Kb and Kw. $\endgroup$
    – Poutnik
    Commented Jul 7, 2021 at 10:03
  • $\begingroup$ @Poutnik I'm not asking if its either A or B; all I did was mention 2 possible explanations I could think of to show my previous work/researh $\endgroup$
    – Vulgar Mechanick
    Commented Jul 7, 2021 at 10:07
  • $\begingroup$ Which one of the 2 (if any) is correct? implies you think both cannot be correct at the same time. If not then I am sorry for misinterpretation, but it could be formulated better. :-) $\endgroup$
    – Poutnik
    Commented Jul 7, 2021 at 10:12

1 Answer 1

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During neutralization of the buffer by a strong base solution, there are ongoing three reversible reactions, maintaining three equilibria:

$$ \begin{align} \ce{H2O + A- &<=> OH- + HA} &\quad K_\mathrm{b} &= \frac{\ce{[OH-][HA]}}{\ce{[A-]}} \tag{1}\\ \ce{H2O &<=> OH- + H+} &\quad K_\mathrm{w} &= \ce{[H+][OH-]} \tag{2}\\ \ce{HA &<=> H+ + A-} &\quad K_\mathrm{a} &= \frac{\ce{[H+][A-]}}{\ce{[HA]}} \tag{3} \end{align} $$

$$K_\mathrm{w} = K_\mathrm{a} \cdot K_\mathrm{b}\tag{4}$$

Therefore, added $\ce{OH-}$ reacts with both $\ce{HA}$ and $\ce{H+}$, reaching again equilibria according to $K_\mathrm{b}$ and $K_\mathrm{w}$.

Additionally, $\ce{HA}$ autodissociation equilibrium makes balance according to $K_\mathrm{a}.$

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  • $\begingroup$ If there are 3 reversible reactions that have to be accounted for, does this mean there's no easy way to calculate the change in pH of the solution upon the addition of a strong base? I'm only familiar with using a single ICE table $\endgroup$
    – Vulgar Mechanick
    Commented Jul 7, 2021 at 17:23
  • $\begingroup$ There is relatively easy way, and even very easy one with few simplifying assumptions. See en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation $\endgroup$
    – Poutnik
    Commented Jul 7, 2021 at 17:28

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