This is what my textbook says. I understand that the strong acid would readily donate a hydrogen ion, but I can't see how all the weak base will readily accept it (at least not to "essential completion").


This is a matter of equilibrium. Let's consider the acid-base reaction of $\ce{HCl}$ and $\ce{NH3}$:

$$\ce{NH3 + HCl <=> NH4+ + Cl-}$$

Which can be broken apart into:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

$$\ce{HCl + H2O <=> H3O+ + Cl-}$$

For these reactions at equilibrium:

$$K_\mathrm{eq}=\ce{\frac{[NH4+][Cl-]}{[NH3][HCl]}} \hspace{1cm} K_\mathrm{b}(\ce{NH3})=\ce{\frac{[NH4+][OH-]}{[NH3]}} \hspace{1cm} K_\mathrm{a}(\ce{HCl})=\ce{\frac{[H3O+][Cl-]}{[HCl]}} $$

$$K_\mathrm{a}(\ce{HCl}) \cdot K_\mathrm{b}(\ce{NH3}) = \ce{\frac{[H3O+][Cl-]}{[HCl]}} \cdot \ce{\frac{[NH4+][OH-]}{[NH3]}}=\ce{\frac{[NH4+][Cl-]}{[NH3][HCl]}} \cdot \ce{[H3O+][OH-]} = K_\mathrm{w}K_\mathrm{eq}$$

$$K_\mathrm{w}K_\mathrm{eq}=K_\mathrm{a}(\ce{HCl}) \cdot K_\mathrm{b}(\ce{NH3}) \Rightarrow$$

$$K_\mathrm{eq}=K_\mathrm{a}(\ce{HCl}) \cdot \frac{K_\mathrm{b}(\ce{NH3})}{K_\mathrm{w}}=\frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{NH4+})} \Rightarrow$$

$$\mathrm{p}K_\mathrm{eq} = -\log \left(\frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{NH4+})} \right) = \log \left(\frac{K_\mathrm{a}(\ce{NH4+})}{K_\mathrm{a}(\ce{HCl})} \right) = \mathrm{p}K_\mathrm{a}(\ce{NH4+}) - \mathrm{p}K_\mathrm{a}(\ce{HCl})=9.24-(-8)=17.24$$

Just from the $\mathrm{p}K_\mathrm{a}$ difference of the acid and conjugate acid of our reaction, we figured out that at equilibrium the products are present in a $10^{17.24}:1$ ratio to the reactants, meaning the reaction has essentially gone to completion.


It is the strong acid that is pushing the weak base to full protonation.

The more formal explanation requires you to be familiar with the ionisation of water $(1)$ and a generic Brønsted acid-base reaction including the acidity constant $(2\text{ – acid})$ and $(3\text{ – base})$.

$$\begin{align}\ce{2 H2O &<<=> H3O+ + OH-}&K_\mathrm{w} &= [\ce{H3O+}][\ce{OH-}] = 10^{-14}\tag{1}\\[0.6em] \ce{HA + H2O &<=> H3O+ + A-}&K_\mathrm{a} &= \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}\tag{2}\\[0.6em] \ce{H2O + B &<=> HB+ + OH-}&K_\mathrm{b} &= \frac{[\ce{HB+}][\ce{OH-}]}{[\ce{B}]}\tag{3}\end{align}$$

The first equation is basically the water-version of the second but without the activity of water in the denominator. Because water is a very weak acid, the equilibrium of $(1)$ lies strongly on the reactant side. If we consider $\ce{B} = \ce{A-}$, i.e. $\ce{B}$ being the conjugate base of $\ce{HA}$ and then add together $(2)$ and $(3)$, we arrive at $(4)$:

$$\require{cancel}\ce{\cancel{\ce{HA}} + 2 H2O + \cancel{\ce{A-}} <=> H3O+ + OH- + \cancel{\ce{HA}} + \cancel{\ce{A-}}} \quad K' = K_\mathrm{a}\times K_\mathrm{b} = K_\mathrm{w}\tag{4}$$

Using the more commonly applied $\mathrm{p}K_\mathrm{a/b}$ values, we see that:

$$\mathrm{p}K_\mathrm{a}(\ce{HA}) + \mathrm{p}K_\mathrm{b}(\ce{A-}) = 14\tag{5}$$

Knowing that strong acids are defined as $\mathrm{p}K_\mathrm{a} < 0$ and weak bases as $\mathrm{p}K_\mathrm{b} >0$, we basically have access to all our parameters; most importantly, the conjugate acid of the weak base must have an acidity constant fulfilling $\mathrm{p}K_\mathrm{a} < 14$ — but at the same time it would still be larger than zero because otherwise the compound would not be notably basic.

Putting this together, we start off with three compounds:

  • a strong acid $\ce{HA}$ with a $\mathrm{p}K_\mathrm{a} < 0$.

  • water, with $\mathrm{p}K_\mathrm{b} = 14$, i.e. $\mathrm{p}K_\mathrm{a}(\ce{H3O+}) = 0$

  • the weak base $\ce{B}$ for which we can say $0 < \mathrm{p}K_\mathrm{b} < 14$ and therefore also $0 < \mathrm{p}K_\mathrm{a} (\ce{HB+}) < 14$

It is clear that the strongest acid — $\ce{HA}$ — will deprotonate practically completely. The question is who gets those protons? We have the choice of giving them to water or the weak base. Whether you now wish to argue with relative basicities (water is less basic than the weak base) or relative acidities of the conjugate acids (hydronium (or oxonium, $\ce{H3O+}$) is more acidic than $\ce{HB+}$) does not matter, in both cases the conclusion is that the weak base acquires all the protons available for thermodynamic reasons.

  • $\begingroup$ This did it for me after grappling with it for so long. So from what I understand, the strong acid will donate its protons to the weak base because the weak base is still stronger than water? $\endgroup$
    – khajiit
    Feb 22 '17 at 21:13
  • $\begingroup$ @khajiit That’s the gist, yes. Because water is such a terrible base. $\endgroup$
    – Jan
    Feb 22 '17 at 22:02

The definition of a weak base as given by Wikipedia is:

In chemistry, a weak base is a chemical base that does not ionize fully in an aqueous solution. As Brønsted–Lowry bases are proton acceptors, a weak base may also be defined as a chemical base in which protonation is incomplete.

Where Brønsted–Lowry is they type of acid-base classification used for your question.

A simple qualitative way to envision what is happening when you add a strong acid to a weak base is that you are "flooding" the unprotonated fraction of the compound with protons, greatly increasing the probability that each of them will become protonated. This is equivalent to saying the reaction of the weak base to it's protonated form essentially goes to completion.


Ugh... Let's not make this hard.

I start off with $n_1$ moles of the free weak base before the reaction and I'll end up with $n_2$ moles of the unprotonated base and $n_3$ moles of the protonated base after the reaction.

So "essential complete” is a significant figures argument. It means that:

$n_3 \approx n1$ and $n_2 \ce{<<} n_1$


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