What actually happens in strong acid- weak base reactions?

Question

I'm going back through my notes from a chemistry class, which I did well in despite not having a very good conceptual understanding, in order to hopefully develop that understanding, but there's something I can't find in the book or online. Why is it that, when solving for the pH of a solution that results from the mixture of a strong acid with a weak base or a weak acid with a strong base, we can treat the acid-base reaction as going to completion before solving any resulting equilibrium problem, or completely ignoring any equilibrium in the pre-mixed solutions.

For a concrete example, when mixing 40.0 mL of 0.50 M ammonia with 25.0 mL of 0.30 M hydrochloric acid, the approach the class would have me use is to first solve the limiting reaction problem as if the reaction goes to completion. Doing so, we see that there are 0.0075 mol of chloride ions, 0.0075 mol of ammonium, and 0.013 mol of ammonia. Then, because we don't have excess hydrogen ions, nor excess hydroxide, nor only neutral ions, we solve the equilibrium problem of ammonium with ammonia.

What I don't understand is how we can ignore the fact that the ammonia solution would have already been in equilibrium when mixed with the hydrochloric acid, and as I understand it should be constantly arriving towards equilibrium as the reaction is taking place.

Alternatively, if there had been excess hydrogen ions, I was told to use the concentration of hydrogen ions from the completed reaction, but that would also be ignoring the fact that the ammonium contributes to the pH (because it again ignores it's equilibrium).

Answer 1

The strategy of major species

You correctly describe the strategy. First, let reactions go to completion (in the direction that makes sense, i.e. weak base and strong acid forms weak acid and spectator ion - never the other way around). Then, check what the major species are and estimate the pH. Finally (and this step is often omitted), see if equilibria involving minor species (such as hydroxide for acidic solutions) need adjustment.

Why does the strategy of major species work?

For the example, the OP already calculated the estimated amount of species: 0.0075 mol of ammonium, and 0.013 mol of ammonia. With a total volume of 65 mL, that comes out as:

$$c_\ce{NH4+} = \pu{0.115M}$$ $$c_\ce{NH3} = \pu{0.20M}$$

From this, we can estimate the pH to be a bit more basic than the $\mathrm{p}K_\mathrm{a}$, i.e. 9.49. Now we can think about the minor species, hydroxide and hydronium. At pH = 9.49, there should be an excess of hydroxide over hydronium, by about 0.00003 M. Dissociation of water does not give an excess, so it has to come from ammonia turning into ammonium:

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

However, when this reaction makes hydroxide, it also changes the ratio of ammonium to ammonia, changing the pH. The only reason we don't keep running in circles, adjusting one equilibrium and then the other, is that this adjustment is tiny compared to the concentration of the major species. It amounts to decreasing the ammonia concentration by 0.00003 M and increasing the ammonium concentration by the same amount. The shift in pH is so small that it makes no difference when the pH is written with appropriate number of significant figures.

Alternate strategy

For the calculators we use, this is a good strategy to deal with multiple equilibria (in your case, that of the weak acid/base pair and autodissociation of water).

If you had an analog computer that adjusts the concentrations based on a "pH slider" and reports out whether all the constraints are met, there would be no reason to do this in multiple steps. Instead, you would just push your pH slider from one extreme to the other and stop when all equations are satisfied.

In your case, you could calculate the ratio of ammonia to ammonium from the pH and the $\mathrm{p}K_\mathrm{a}$, and the hydroxide concentration from the autodissociation constant of water. Then, you vary the pH until the charge balance (chloride plus hydroxide has to match hydronium plus ammonium) is achieved. It is a fun exercise to program this into a spread sheet.

What actually happens?

All reactions (any of the acids - water, hydronium or ammonium - reacting with any of the bases - water, ammonia or hydroxide) go on at the same time. The further from equilibrium a reaction is, the faster is the net change. Once all the reactions are at equilibrium, there is no more net change. The details depend on the kinetics but are not relevant to the equilibrium state that is finally reached.

Answer 2

Given the following problem:

25.0 mL of 0.30 M hydrochloric acid is added to 40.0 mL of 0.50 M ammonia. Calculate the pH, and concentrations of $\ce{NH3 \text{and} NH4+}$.

To solve such problem you must make some assumptions. Some of which are good, some ok, and some bad. In math $\pi$ has been calculated to 50 trillion digits. Chemistry generally works with 2-4 significant figures.

Note that the concentrations of both hydrochloric acid and ammonia have only been given to 2 significant figures.

Assumptions:

1. The final concentrations of $\ce{NH3, NH4+ \text{and} H+/OH-}$ at equilibrium are independent of how the solution was prepared.

The equilibrium doesn't depend on how the solution was mixed. So regardless of how 65 ml of solution containing the equivalent of 7.5 millimoles of hydrochloric acid, and 20 millimoles of ammonia was made, it will have the same equilibrium.

Also $z$ millimoles of ammonium chloride could be used ($z \le 7.5$). Then $7.5-z$ millimoles of hydrochloric acid and $20-x$ moles of ammonia would be needed in the mixture.

2. That the volumes of the hydrochloric acid and ammonia solutions are additive.

Certainly good to 2 significant figures, 3 might be good, but 4 is probably an overreach.

3. That concentrations can be used instead of activity coefficients.

This is definitely a bad assumption for such concentrated solutions.

4. That the reaction $\ce{H+ + NH3 -> NH4+}$ is quantitative.

This is ok for 4 significant figures and the given concentrations.

5. Expanding 4, also assume that all the $\ce{NH4+}$ is due to the reaction with hydrochloric acid.

This assumption is good to 2 significant figures, probably 3, but 4 may be a issue. The issue is that there is also the autoionization of water which shifts the equilibrium $\ce{NH3 + H2O <=> NH4+ + OH-}$

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

---

There is really nothing that can be done with assumptions 2 and 3 without experimental data and/or some sort of functions for the data. So the only thing that can be done is to note the assumptions in the solution.

From Wikipedia the $K_\mathrm{b}$ for ammonium hydroxide is $1.8\times10^{-5}$ ($\mathrm{p}K_\mathrm{b} = 4.7447$, $\mathrm{p}K_\mathrm{a} = 9.2552$).

---

Solution 1 Ignoring the autoionization of water yields the following charge balance equation:

$\ce{[NH4+] \approx [Cl-]}$

Thus 7.5 millimoles of $\ce{NH4+}$ and 12.5 millimoles of $\ce{NH3}$.

$$\ce{[NH4+] = \dfrac{\pu{7.5 millimoles}}{\pu{65 mL}}} = \pu{0.1153 M}$$

$$\ce{[NH3] = \dfrac{\pu{12.5 millimoles}}{\pu{65 mL}}} =\pu{0.1923 M}$$

Using the hammer known as the Henderson–Hasselbalch equation:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\left(\dfrac{\mathrm{[base]}}{\mathrm{[acid]}}\right) = 9.2552 + \log\left(\dfrac{\pu{0.1923 M}}{\pu{0.1153 M}}\right) = 9.4773 $$

Now rounding to two significant figures:

$$\ce{[NH4+]} = \pu{0.12 M}$$

$$\ce{[NH3]} = \pu{0.19 M}$$

$$\mathrm{pH} = 9.47$$

Solution 2 Considering the autoionization of water yields the following charge balance equation:

$\ce{[NH4+] + [H+] = [Cl-] + [OH-]}$

To determine if the first solution is adequate, do two checks using the values from the first solution:

• Is $\ce{[NH4+] \gg [H+]}\ $?

$$\pu{0.12 M}\ \ce{NH4+} \gg \pu{3.4\times10^{-10} M}\ \ce{H+}$$

So this assumption is good.

• Is $\ce{[Cl-] \gg [OH-]}\ $?

$$\pu{0.19 M}\ \ce{Cl-} \gg \pu{3.0\times10^{-5} M}\ \ce{OH-}$$

So this assumption is good to 2 significant figures, but barely good for 4 significant figures.

Thus the first solution is adequate