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I have looked at several sources including Khan Academy, ChemCollective and the answer to a similar question: Can someone please explain buffers to me?

However, it is not clear whether buffers can only protect against the opposite species. I.e. a weak acid can only protect against the pH change caused by strong bases, or if it can also protect against the pH change caused by strong acids.

This is my thought:

If we have $\ce {H_2CO_3_{(aq)} + H_2O_{(l)}<=>H_CO^{-}_3_{(aq)} + H_3O^+_{(aq)}}$

Then we add a strong acid $\ce{HCl + H_2O-> H_3O^+ + Cl^-}$

Then we have $x$ amount of protons in the solution and for every one of the weak acid's conjugate base, there is a proton. So, to me it appears that there is no way for the weak acid to act as a buffer for a strong acid. Is this correct or am I missing something?

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    $\begingroup$ Please consider that carbonic acid is actually not very stable in aqueous solution. $$\ce{H2CO3 <=>[\ce{H2O}] H2O + CO2}$$ For a buffer solutions it is better to use something along the lines of $$\ce{KHCO3 + H2O <=> KCO3^{-} + H3+O}.$$ $\endgroup$
    – Martin - マーチン ♦
    Jun 12, 2014 at 5:51
  • $\begingroup$ @Martin I chose this simply because it is the pH buffer in our blood. Ref (chemistry.wustl.edu/~edudev/LabTutorials/Buffer/Buffer.html) $\endgroup$
    – Klik
    Jun 12, 2014 at 5:54
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    $\begingroup$ Actually, all 3 ionization states of carbonic acid are used to buffer our blood. Carbonic acid is unstable in solution and that it can be readily converted to carbon dioxide allows us to, effectively, exhale acid and maintain the buffer system. $\endgroup$
    – canadianer
    Jun 12, 2014 at 6:00

2 Answers 2

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Think of it in terms of the equilibrium constant:

$$K_{c}\; =\; \frac{[\ce{H3O+}][\ce{HCO3-}]}{[\ce{H2CO3}]}$$ Let's assume for a moment that there are 2 mols of $\ce{H3O+}$, 2 mols of $\ce{HCO3-}$, and 1 mol of $\ce{H2CO3}$. In our pretend scenario, $K_c$ is 4 (however in reality the $K_c$ is much lower and is about $2.5 \cdot 10^{-4}$). When you add an acid, the $\ce{H3O+}$ concentration rises. Lets assume there are now 9 mols of $\ce{H3O+}$. So now the reaction quotient is

$$Q\; =\; \frac{9\cdot 2}{1} = 18$$

Because $Q > K$ the reaction has to shift left in order to compensate and go back to the equilibrium state. If 1 mol of $\ce{H3O+}$ recombines with 1 mol of $\ce{HCO3-}$ to give back 1 mols of $\ce{H2CO3}$ now you have

$$Q\; =\; \frac{8\cdot 1}{2} = 4$$

and because $Q = K$ the reaction is now at equilibrium.

So while the total number of mols of $\ce{H3O+}$ did increase, the effect of it was dampened by the buffer (as it only went up to 8 mols instead of 9 mols).

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If you just add straight carbonic acid to water, the $\mathrm{p}\ce{H}$ will drop as it is ionized (as represented by your equation). Since it is a weak acid it won't ionize completely and you'll be left with carbonic acid, bicarbonate and hydronium in solution at equilibrium. Now, if you increase the concentration of $\ce{H+}$ (such as by adding an acid), Le Chatelier's principle dictates that the equilibrium will be disturbed and the system will move in the direction that will restore the equilibrium (in this case to the left, as protons react with bicarbonate to produce carbonic acid). Because the new protons are not free in solution, the pH will not drastically change until the bicarbonate is exhausted. Thus the solution is buffered.

Mathematically, the acid dissociation constant, which is unchanging for a specific species by definition, is given by $K_a=\frac{[\ce{HCO3^{-}}][\ce{H+}]}{[\ce{H2CO3}]}$. If you add more $\ce{H+}$, the ratio of $\ce{HCO3^-}$ to $\ce{H2CO3}$ must decrease for the system to regain equilibrium

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  • $\begingroup$ So you're saying a weak acid can act as a pH buffer in a solution when a strong acid is introduced? The reason I don't understand is because for every conjugate base of the weak acid, there is a proton that it has dissociated from. [Conjugate base] = [H+]. When the strong acid is introduced (a supply of H+). The weak acid's conjugate base may react with the excess [H+], but for every conjugate base that reacts with an H+, Le Chatelier's Principle would require the weak acid to produce more conjugate base and thus more H+. In effect, I don't see how H+ is reduced. $\endgroup$
    – Klik
    Jun 12, 2014 at 4:16
  • $\begingroup$ No, because when the conjugate base is protonated, you get the original acid again. $\endgroup$
    – Dissenter
    Jun 12, 2014 at 5:18
  • $\begingroup$ @Klik I think you applied Le Chatelier's Principle incorrectly. The equilibrium shifts in the other direction. Because you added more [H+] in order for the equilibrium constant to remain the same the reaction must shift left and produce $\ce{H2CO3}$. $\endgroup$
    – 1110101001
    Jun 12, 2014 at 5:18
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    $\begingroup$ I'm saying that adding a weak acid to water produces it's conjugate base and that that system will buffer pH changes in either direction. If you increase the concentration of H+ (such as by adding an acid), Le Chatelier's principle dictates that the equilibrium will be disturbed and the system will move in the direction that will restore the equilibrium (in this case to the left). The acid dissociation constant, which is unchanging for a specific species by definition, is given by Ka=[A-][H+]/[HA]. If you add more H+, the ratio of A- to HA must decrease for the system to regain equilibrium. $\endgroup$
    – canadianer
    Jun 12, 2014 at 5:25
  • $\begingroup$ @canadianer I understand what you're saying now. Could you include the details you just explained to me in your answer? I will mark it as correct. Thanks for the explanation. $\endgroup$
    – Klik
    Jun 12, 2014 at 5:36

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