I learned that

a) the conjugate base to a weak acid is a strong base (and vice versa).

b) a buffer consists of a weak acid (base) and its conjugate base (acid).

However, this explanation of buffers says the following:

"[...] a weak acid is one that only rarely dissociates in water [...]. Likewise, since the conjugate base is a weak base, [...]"

which seems to stand in conflict with my assumption a) above. So, what is correct?

Furthermore, if b) is correct, isn't any solution of a weak acid solution a buffer, since any weak acid in water makes an equilibrium of the form

$HA \text{ (weak acid)} \leftrightarrow H^+ + A^- \text{ (conjugate base)}$

and is thus a solution of a weak acid and its base?


2 Answers 2


Your statement a) isn't always true.

Water dissociation is represented by:

$$\ce{H2O + H2O <-> H3O+ + OH-}$$

$$ K_\mathrm w=[\ce{H3O+}]\cdot [\ce{OH-}] = 1\times 10^{-14}\ (\textrm{at}\ 25^\circ ~\mathrm C) $$

Note 1: We don't write the $\ce{H_2O}$ activity, since it can usually be rounded to 1 and the ions activities can be rounded to their concentrations.

Note 2: This value was obtained experimentally, considering the concentration of $\ce{H_3O^+}$ and $\ce{OH^-}$ in the medium was the same and measuring the ionization of water.

On a weak acid dissociation:

$$\ce{HA + H2O <-> A- + H3O+}$$

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

On its conjugate base:

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

$$K_\mathrm b=\frac{[\ce{HA}]\cdot [\ce{OH^-}]}{[\ce{A^-}]}$$

If we 'add' both reactions we expect that both equilibriums will happen in the mixture, so we have:

$$\ce{HA + H2O <-> A- + H3O+}\ \ \ \ (K_\mathrm a)$$


$$\ce{A- + H2O <-> HA + OH-}\ \ \ \ (K_\mathrm b)$$


$$\ce{H2O + H2O <-> H3O+ + OH-}\ \ \ \ (K_\mathrm w)$$

$$K_\mathrm b\cdot K_\mathrm a=[\ce{H3O+}] \cdot [\ce{OH-}] \cdot ([\ce{A-}]\cdot [\ce{HA}])/([\ce{A-}] \cdot [\ce{HA}]) = [\ce{H_3O^+}]\cdot [\ce{OH^⁻}] = K_\mathrm w$$

$$K_\mathrm b=K_\mathrm w/K_\mathrm a~~~~~~ K_\mathrm b=10^{-14}/K_\mathrm a\ \ \textrm{at}\ 25^\circ ~\mathrm C$$

Another way to write this:

\begin{align}-\log(K_\mathrm b)&=-\log(10^{-14}K_\mathrm a) \\ \implies \mathrm pK_\mathrm b &= -\log(10^{⁻14}) - (-\log(K_\mathrm a))\\ \implies \mathrm pK_\mathrm a+\mathrm pK_\mathrm b &=14\;.\end{align}

This is the relation between a conjugate base strength and its acid strength. A very strong acid has a weak conjugate base, but a weak acid doesn't necessarily have a very strong base. Say you have a $\mathrm pK_\mathrm a=5$, which is a weak acid, with $K_\mathrm a=1\times 10^{-5}$. The conjugate base would have a $\mathrm pK_\mathrm b=14-5=9$ or a $K_\mathrm b=1\times 10^{-9}$, which is not a strong base.

However, if we have a strong acid, like $\ce{HCl}$ with a $\mathrm pK_\mathrm a=-6.3$ and $K_\mathrm a=10^{6.3}$. Its conjugate base would have a $\mathrm pK_\mathrm b=14-(-6.3)=20.3$ and $K_\mathrm b=10^{-20.3}$ which is a really weak base.

Hopefully I didn't make it even more confusing for you! It's mostly an reaction equilibrium issue.


A strong acid (or base) forms a weak conjugate base (or acid). This is correct. By saying that $\ce{HA}$ is a weak acid and $\ce{A-}$ is a weak conjugate base they mean:

  • Acid shouldn't be too strong - don't use $\ce{NaCl + HCl }$
  • Conjugated base shouldn't be too strong - don't use $\ce{EtOH/NaOEt }$
  • Use a "not-strong" acid (weak acid) whose conjugate base is also "not-strong" (weak conjugate base). For example, $\ce{H3CCOOH}$ is a relatively weak acid. But $\ce{H3CCOONa}$ is also a relatively weak base. Choose such acids to make a buffer.

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