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?


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.

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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