How can adding a strong base to a weak base be the same as a strong base in solution problem?

Question

In professor's lecture on acid-base titration, professor says (at this point of the video, link should start about the time she says it)

"adding a strong base to a weak base should be treated as a strong base problem".

I disagree with it. I'll treat below the analog case of strong acid introduced in weak acid solution.

Since $K_{eq}$ of the overall reaction (of the strong acid with the weak acid) is equal to $K_{a_1} K_{a_2}$. And let's take for $K_{a_1}=10^3$ (strong acid, correpsonding to a $pK_{a_1}=-3$) and and $K_{a_2}= 10^{-5}$ (weak acid, corresponding to a $pK_{a_2}=5$ ). The resulting $K_{eq}$ will be equal to $10^{-2}$ i.e. there won't be tons of hydroxonium ions in the resulting solution. In other words it is not like a strong base in water problem.

I know there should be something wrong with my reasonning. I don't understand the professor sentence when it comes to strong base in weak base (or analogly strong acid in weak acid).

Answer 1

It might (or might not) be helpful to derive this result irectly from the mathematical equations.

Say the strong base is $\ce{KOH}$ and the weak base is $\ce{NH3}$. There are four known quantities: the amounts of the two bases that we add to the solution, and their $K_b$ values. I'll assume for the moment that we added 0.1 mol of $\ce{NH3}$ and 0.1 mol of $\ce{KOH}$ to a liter of water. The $K_b$ values will be $K_{strong}$ and $K_{weak}$, with $K_{strong}$ being large (let's say 1000 although it's not actually that high for $\ce{KOH}$) and $K_{weak}$ being small, like $10^{-5}$ for ammonia, or even less for a weaker base.

Then there are a total of five unknown quantities: $\ce{[OH-]}, \ce{[K+]}, \ce{[KOH]}, \ce{[NH3]}, \ce{[NH4+]}$, all in moles per liter. We really want to know $\ce{[OH-]}$. If we were adding only the strong base, the final $\ce{[OH-]}$ would be 0.1, and the claim is that even if the weak base is added too, it will still be very close to that value.

We have the following equations, with brief explanations of where they come from:

• Dissosciation of the strong base: $$\frac{\ce{[K+]}\ce{[OH-]}}{\ce{[KOH]}} = K_{strong}\tag{1}$$
• Reaction of the weak base with water: $$\frac{\ce{[NH4+]}\ce{[OH-]}}{\ce{[NH3]}} = K_{weak}\tag{2}$$
• Conservation of $\ce{K}$: $$\ce{[K+]} + \ce{[KOH]} = 0.1\tag{3}$$
• Conservation of $\ce{NH3}$: $$\ce{[NH4+]} + \ce{[NH3]} = 0.1\tag{4}$$
• Charge neutrality: $$\ce{[K+]} + \ce{[NH4+]} = \ce{[OH-]}\tag{5}$$

Five equations and five unknowns. Nice.

One way to solve these equations is to substitute (3) into (1) and (4) into (2), eliminating two variables and giving the following equations:

$$\ce{[K+]} = \frac{0.1}{1+K_{strong}^{-1}\ce{[OH-]}} \tag{1a}$$ $$\ce{[NH4+]} = \frac{0.1}{1+K_{weak}^{-1}\ce{[OH-]}} \tag{2a}$$

We can then substitute (1a) and (2a) into (5) to get:

$$ \frac{0.1}{1+K_{strong}^{-1}\ce{[OH-]}} + \frac{0.1}{1+K_{weak}^{-1}\ce{[OH-]}} = \ce{[OH-]}.\tag{5a}$$

Now, this can be solved exactly (it's a cubic equation of one variable), and if you want, you can plug in the real values of $K_{strong}$ and $K_{weak}$ for this problem and do so. But it's more instructive to analyze it qualitatively. The first thing to observe is that the left side is decreasing and the right side is increasing. So there is going to be at most one solution.

Next, we know $K_{strong}$ is large and $K_{weak}$ is very small. Therefore $K_{strong}^{-1}$ is small and $K_{weak}^{-1}$ is very large. This means that the denominator of the first fraction is going to be close to 1, while the denominator of the second fraction is going to be very large. The left-hand side is therefore approximately $$\frac{0.1}{1+\mathrm{small}} + \frac{0.1}{\mathrm{large}} \approx 0.1$$ and we deduce that $\ce{[OH-]}$ should thus be approximately $0.1$, just as if we were adding the strong base by itself.

Answer 2

You have a weak base, like ammonia in water. Then you add a strong base, like sodium hydroxide. This is the equilibrium of ammonia in water

$NH_3 + H_2O \rightleftarrows NH_4^+ + OH^- $

Now the sodium hydroxide reaction in water:

$NaOH \rightarrow Na^+ + OH^-$

These are two different chemical reactions, with 2 independent equilibrium constants. You cannot multiply them, like you did.

Returning to the mixture, the sodium hydroxide is a strong base. This means that it is completely dissociated in ions. But the hydroxide ions are also the product of the ammonia equilibrium. Then for the Le Chatelier's Principle, the reaction equilibrium is shifted to the left. Practically speaking, concentration of the hydroxide ions from $NH_3$ is negiglible.