Consider a buffer solution containing weak acid $\ce{CH3COOH}$ and its salt $\ce{CH3COONa}$: \begin{align} \ce{CH3COOH &<=> CH3COO- + H+} \tag{1}\\ \ce{CH3COONa &-> CH3COO- + Na+} \tag{2} \end{align}

These two reactions take place and now we are majorly left with $\ce{H+}$, $\ce{CH3COOH}$, $\ce{CH3COO-}$, $\ce{Na+}$ in the solution. Since, $\ce{Na+}$ is a spectator ion, we can ignore it.

We know that buffers resist change in pH on adding small quantities of $\ce{H+}$ and $\ce{OH-}$. Let us add $\ce{H^+}$ to this solution.

A buffer solution must contain some things that could remove these $\ce{H+}$ ions, otherwise the pH would change. As we add $\ce{H+}$ to this buffer solution, it will react with $\ce{CH3COO-}$ to form $\ce{CH3COOH}$.

But here's the catch.

$\ce{CH3COO-}$ is getting removed and $\ce{CH3COOH}$ is getting formed. According to Le Chatelier's principle (LCP), the equilibrium should shift towards in the forward reaction (in reaction (1)). Shifting of equilibrium in the forward reaction should increase the concentration of $\ce{H+}$ and hence decrease the pH of the solution.

Now, one might say that $\ce{CH3COOH}$ is a weak acid and it would not decompose much. So, we can actually ignore the amount of $\ce{H+}$ so generated.

Let's "assume" that's true for a while and move ahead.

Let us add $\ce{OH-}$ ions to the solution.

Remember there's $\ce{H+}$, $\ce{CH3COOH}$, $\ce{CH3COO-}$, $\ce{Na+}$ present in the solution. $\ce{OH-}$ can react with either (a) $\ce{H+}$, or (b) $\mathrm{CH_3COOH}$.

Here comes the tricky part.

Let us only consider hydroxide ions reacting with $\ce{H+}$

If $\ce{OH-}$ reacts with $\ce{H+}$, water will be formed.

Here although hydroxide ions are getting removed, $\ce{H+}$ are also being removed from the solution. Now, because of the removal of $\ce{H+}$ shouldn't the solution become less acidic and the pH should increase.

Now, some might say as soon as $\ce{H+}$ is removed, the equilibrium in the first reaction shift towards the right.

But wait, didn't we just ignore that (in the case when we added $\ce{H+}$ in the solution)?

Can anyone explain why is there a contradiction?

P. S.: I have ignored the $\ce{H+}$ and $\ce{OH-}$ from water because it is irrelevant to the discussion.

  • 4
    $\begingroup$ pH will change all right, only way less than it would change in pure water. $\endgroup$ Commented Aug 12, 2017 at 16:40
  • 1
    $\begingroup$ we can't ignore the forward reaction in the second case as the net equilibrium(of water formation and decompostion of CH3COOH) would be around 10^10.(10^15 of water and 10^(-5) of acetic acid) $\endgroup$
    – Ayushmaan
    Commented Aug 12, 2017 at 16:51
  • $\begingroup$ @Ayushmaan and we can ignore it in the first casE? $\endgroup$
    – Arishta
    Commented Aug 12, 2017 at 16:52
  • $\begingroup$ yes, because in the first case the equilibrium constant would only be 1. ( 10^5*10^(-5)). So, if initial concentration of acetic acid is large in comparison to H+ we have put, the pH would be approx. same. $\endgroup$
    – Ayushmaan
    Commented Aug 12, 2017 at 16:59

2 Answers 2


Having problems with that concept myself, above answers seem mighty complicated, this is how I think it works (warning- only my understanding, could well be wrong)

Point One: You have a lot of $\ce{CH3COOH}$ (not much decomposition) and a lot of $\ce{CH3COO−}$ (from the salt) and very little $\ce{H+}$

Point Two: pH works on the amount of $\ce{OH-}$ and $\ce{H+}$ in the system, there are usually only very very small amounts (when looking at weak acids and bases) present compared to the reactant (in water, there is about 500 million molecules to ever one that dissociates) So the point I'm making is adding a few $\ce{OH-}$ or $\ce{H+}$ ions will have a big impact on the pH.

Conclusion drawn: As you have so much $\ce{CH3COO−}$ to begin with, removing a very very small amount by adding a few $\ce{H+}$ ions and forming a very very small amount of $\ce{CH3COOH}$, is not going to have a noticeable impact on the concentrations, in comparison to the amount already in the solution (both $\ce{CH3COO−}$ and $\ce{CH3COOH}$). You could even say that it is a negligible amount and can be ignored, therefore can say this will not have an effect on the equilibrium.

As the $\ce{H+}$ is removed, the pH is not changed; however as there is so little $\ce{H+}$ ions to begin with, not removing them will have an impact on the pH (see point two).

As already stated, there are so little $\ce{H+}$ ions in the system, removing them (by adding $\ce{OH-}$) will have an impact on equilibrium, you cannot say the amount of $\ce{H+}$ removed is negligible, you cannot ignore Le Chatelier's principle, the equilibrium will shift right to replace the lost $\ce{H+}$.

Note: I'm an A-level student, don't take this as 100% right or even 1% right, this, at best is a calculated guess as to why equilibrium can be ignored when you add $\ce{H+}$ ions and equilibrium can not be ignored when you add $\ce{OH-}$.


The characteristic feature of a buffer is that the equilibrium is balanced at an inflection point with $\mathrm{HA}$ and $\mathrm{A^-}$ set approximately equal. The $[\mathrm{H}^+]$ concentration changes very little here with a change in the ratio $\mathrm{[A^-]/[HA]}$.

Let's look at an imaginary case, with numbers: $K_{\mathrm{a}}$ of acid $\mathrm{HA} = 10^{-6}$. Then

$~~~~~~ K_{\mathrm{a}} = 10^{-6} = \mathrm{\dfrac{[H^+][A^-]}{[HA]}}$

For a 1 M solution of $\mathrm{HA}$, $\mathrm{[H^+] = [A^-] = 0.001~}M$. The pH will be 3.0.

These are very small amounts, reflecting the fact that weak acids are weakly ionized. If you add 0.1 mole of strong acid to 1 L of solution, the pH will change to ~1.0, a significant change.

If you add 0.1 mole of strong base, the pH increases:

$~~~~~~ K_{\mathrm{a}} = 10^{-6} = \mathrm{\dfrac{[H^+][\sim 0.1]}{[\sim 0.9]}} \;\; \therefore \;\; [\mathrm{H}^+] = 9.1\times 10^{-6}$ and pH goes to ~5.04.

If we make 1 L buffer solution with HA = 0.5 M plus 0.5 M of an alkali salt of HA,

$~~~~~~ K_{\mathrm{a}} = 10^{-6} = \mathrm{\dfrac{[H^+][\sim 0.5]}{[\sim 0.5]}} \;\; \therefore \;\; [\mathrm{H}^+] = 1 \times 10^{-6}$ and pH = 6.

Now add 0.1 mole of strong acid ($\mathrm{A^-}$ is consumed, $\mathrm{HA}$ increases):

$~~~~~~ K_{\mathrm{a}} = 10^{-6} = \mathrm{\dfrac{[H^+][\sim 0.4]}{[\sim 0.6]}} \;\; \therefore \;\; [\mathrm{H}^+] = 1.5\times 10^{-6}$ and pH drops to 5.82.

This is a relatively small change in pH.

If you add 0.1 mole of strong base, $\mathrm{HA}$ is consumed and $\mathrm{A^-}$ increases:

$~~~~~~ K_{\mathrm{a}} = 10^{-6} = \mathrm{\dfrac{[H^+][\sim 0.6]}{[\sim 0.4]}} \;\; \therefore \;\; [\mathrm{H}^+] = 0.66\times 10^{-6}$ and pH rises to 6.17.

Unbuffered, the range of pH change from addition of 0.1 M strong acid or strong base is ~4 pH units around the starting pH of 3; buffered, the range is only 0.35 pH units around the starting pH of 6. Adding less of the strong acid or base produces an even smaller change in the pH of the buffered solution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.