This is what I understand so far:

The concept of a buffer is to minimize any swings in pH. In order to do this you need to create a solution with both acidic and basic compounds.

To create one, you start with an acid: $\ce{HA + H2O -> A- + H3O+}$

[My first question is why for buffers the acid needs to be weak.]

The buffer is then created by adding in a salt containing the anion of the base which allows for both the acid and its conjugate base to exist in the solution.

Here, why doesn't the conjugate base react with the acid? Second, why does this help minimize pH changes. I see that since that you have a weak acid the conjugate base will be strong and can react with any acid added. However, if you add base what is going to react with it? Isn't the acid too weak to do so? Third, what happens to the $\ce{H3O+}$ that exists in the solution?


1 Answer 1


Writing out the equilibrium expression for a weak acid in solution:

$$[\ce{H3O+}] = K_\mathrm a\frac{[\ce{HA}]}{[\ce{A-}]}$$

Buffers need to keep the concentration of the weak acid to its conjugate base similar in order for it be effective.

Let’s say you add a small amount of base. Some of the $\ce{HA}$ molecules are converted into $\ce{A-}$ ions. The ratio $\frac{[\ce{HA}]}{[\ce{A-}]}$ changes but not by a large amount.

If we add a small amount of acid, we generate more $\ce{HA}$. Again, the ratio $\frac{[\ce{HA}]}{[\ce{A-}]}$ changes but not by a large amount.

Why do buffers for the acid needs to be weak?

If you look at the expression, the concentration of hydronium ions is directly related to the ratio of the conjugate acid and its base. If you take a strong acid for instance, hydrochloric, it will almost completely deprotonate in water. Thus the ratio of $\frac{[\ce{HA}]}{[\ce{A-}]}$ is close to 0 and is negligible, and we use the $K_\mathrm a$ to directly find the concentration of hydronium ions. As strong acids have very large $K_\mathrm a$ values, we can directly take the concentration and assume that will be the concentration of our hydronium ions.

Why doesn’t the conjugate base react with the acid?

It depends on the acid. In buffer solutions, competition arises between acid and the conjugate base. For instance, if the $K_\mathrm a$ is large relative to $10^{-7}$, the acid ionization will proceed and you will have an equilibrium with the conjugate base reacting with its acid resulting in no buffer action.

If however, the tendency of the acid in donating hydrogen ions, say, to water, is outcompeted by the tendency of the base to accept those hydrogen ions from water, this net reaction produces $\ce{OH-}$ ions, the $K_\mathrm a$ of this acid-base pair is less than $10^{-7}$, and we use $K_\mathrm b$ to find equilibrium.

Derivation of the Henderson–Hasselbalch equation:

Let’s use acetic acid and water as an example:

$$\ce{AcOH + H2O <=> AcO- + H3O+} $$

$$ K_\mathrm a = \frac{[\ce{AcO-}][\ce{H3O+}]}{[\ce{AcOH}]}$$

Taking the log of both sides:

$$ \log(K_\mathrm a) = \log\left(\frac{[\ce{AcO-}][\ce{H3O+}]}{[\ce{AcOH}]}\right)$$

Using the law of logs where $\log(xy) = \log(x) + \log(y) $:

\begin{aligned} \log\left(K_\mathrm a\right) &= \log\left(\frac{[\ce{AcO-}]}{[\ce{AcOH}]}\right) + \log([\ce{H3O+}])\\ -\log([\ce{H3O+}]) &= \log\left([\ce{AcO-}][\ce{AcOH}]\right) - \log(K_\mathrm a) \\ \mathrm{pH} &= \log\left(\frac{[\ce{AcO-}]}{[\ce{AcOH}]}\right) + \mathrm{p}K_\mathrm a \\ \mathrm{pH} &= \mathrm{p}K_\mathrm a + \log\left(\frac{[\ce{B}]}{[\ce{A}]}\right) \\ \end{aligned}

With buffer systems, you can calculate the $\mathrm{pH}$ of any buffer solution. The Henderson–Hasselbalch equation has several caveats however.

  1. You need have both the acid and its conjugate base in solution.
  2. The extent of ionization must be small enough so that the concentrations of hydronium and hydroxide ions are small relative $[\ce{HA}]_0$ and $[\ce{A-}]_0$.
  • $\begingroup$ When you have a buffered solution and add a base, why doesn't the base react with the H3O+ ion (which would change the pH)? Why does it prefer to react with the acid instead? $\endgroup$
    – 1110101001
    Mar 13, 2014 at 6:12
  • $\begingroup$ @user2612743: The base would most probably react with $H_3O^+$ as you suggest, but very soon (nanoseconds at most), the acid will "create a new" $H_3O^+$. It is just equilibrium in zillions of molecules. $\endgroup$
    – ssavec
    Mar 13, 2014 at 7:22
  • $\begingroup$ For the part about acids not reacting with conjugate base, is it because the only way the react is $\ce{HA + A- -> A- + HA}$ resulting in no net change of concentration? $\endgroup$
    – 1110101001
    Mar 13, 2014 at 18:16
  • $\begingroup$ And why must only weak acids be used for buffers? Why can't you use a strong one? $\endgroup$
    – 1110101001
    Mar 13, 2014 at 18:16
  • $\begingroup$ @user2612743 I readdressed those issues and clarified them in my post. $\endgroup$
    – user3735
    Mar 13, 2014 at 22:37

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