# In a neutralization reaction between a strong acid and weak base, can I assume the weak base is totally ionized?

Here's a Chemistry question from the infamously terrible Nelson textbook.

10.0 mL of $\ce{NH3(aq)}$, with a concentration of $\mathrm{5.70\dot\ 10^{-2}}$ mol/L, is titrated to the endpoint (by which I assume they mean equivalence point) with $\mathrm{2.85}$ mL of $\ce{HBr}$ solution. What is the concentration of the $\ce{HBr}$ solution?

The answer they give is $\mathrm{0.200}$ mol/L which assumes that all of the moles of $\ce{NH3}$ reacts with the acid. I am unsure of this because I thought weak bases only ionized partially, and the amount they ionize has to be gleaned from the $\mathrm{K_b}$ table.

Why, if it is at all, correct to assume all of the base reacts?

• The magic word is equilibrium. – Zhe Jan 31 '17 at 3:42

You're never going to puzzle this out without some guidance so here goes...

Why, if it is at all, correct to assume all of the base reacts?

All of the base will never react no matter how much acid you add. If you setup the equilibrium equation you can solve for $\ce{[NH3]}$ for any given pH. Thus there will always be some small amount of $\ce{NH3}$ in an acidic solution.

The gist is that there is $5.70\cdot10^{-4}$ moles of base. If you add "exactly" that much acid then the necessary assumption is that the acid balances the base and the solution is neutralized.

One other thing to consider in all of this is the significant figures. So +/- $1\cdot10^{-6}$ moles of acid doesn't make any real difference. That is enough to swing the final pH from 6 to 8.

Just like for the acid if $\ce{[NH3]} < 1\cdot10^{-6}$ then the free ammonia is negligible to the amount of $\ce{[NH4+]}$. Thus, considering the significant figures, the assumption that all of the ammonia was converted is reasonable, even though it is not actually absolutely true.

Here follows an alternative answer, but only in words.

Dissolving ammonia in water will establish an equilibrium between ammonia and ammonium ions determined by the concentration and the equilibrium constant. The pH of such a solution will be alkaline. Suppose the pH is 10.5. By the titration with the strong acid HBr, all ammonia should be considered to be transformed to ammonium ions at the equivalence point. Since we have added a strong acid to the ammonia solution, the pH will drop to about 5 - 6.

The interesting part is what you get at the equivalence point.

The equivalence point will be exactly equal to dissolving pure ammonium bromide in water in the same concentration as you have at the equivalence point. What will the outcome be of doing such a thing?

Well, the ammonium ions of ammonium bromide will react with water and equilibrate. Ammonia will be formed in equilibrium with ammonium ions This is similar to what we had at the starting point, when ammonia reacted with water. The big difference is that the pH is not the same. A new equilibrium will be established. At the new (lower) pH the relative amount of ammonia will be much lower than at the starting point, i.e. before the titration was initiated.

So, even if you have transformed all initial ammonia to ammonium ions by the titration, there will be ammonia present at the equivalence point because of the new equilibrium established at the lower pH.

Yes. Assume that at the equivalent point, all of the base and all of the acid are "neutralized". That is there is an equivalent amount of acid and base present at the equivalence point, by definition. If you have not worked through the logic of the weak/strong acid/base reactions (WA+WB, WA+SB, SA+WB, & SA+SB ) you should. here's a fairly simple (actually somewhat oversimplified -- NH3 not NH4OH is the principle electrically neutral species) http://www.tutorvista.com/content/chemistry/chemistry-iii/ionic-equilibrium/strong-acid-and-weak-base.php. For a WA+SB salt, its solution (in water) will be basic (pH > 7). For a SA+WB salt such as ammonium hydroxide + hydrobromic acid, the pH of salt solutions will be acidic (pH < 7). Meaning that the titration of NH3 with HBr will have a pH < 7 at the equivalence point. (For a volatile salt such as this there are other potential issues such as does the solid salt decompose? does either the Br- or the NH3 volatilize from solution? BUT ignore these fine points, they're not really relevant to your question) In fact you can calculate the pH, when given the pKa of both (or pKb). Look at it this way. Some of the NH3 is ionized to NH4 and OH- but adding H+ will neutralize all of the OH (in excess of pH 7) allowing more NH3 to ionize. (Pushing the equilibrium toward the right.) The acid pushes the pH down. Meaning the solution will be "more acidic" than what it would be with a strong base. Even at high pH (> 7) there is more H+ than expected (but this is not saying that there's more H+ than OH- above pH 7(!), just there's more than "expected" (for a stong base)). (This is also not saying that at a specific basic pH, that there is more H+ than what is being measure, rather it is saying that the pH is lower than what would be expected given the concentration of Base+ ions present, since WB exists both as B and B+ in basic solution.