# Video: Calculating the pH of Weak Bases from the Acid Dissociation Constant K_b

Which base in the table is most appropriate for the preparation of a buffer solution with a pH of 9.28? [A] (CH₃)₂NH [B] (CH₃)₃N [C] CH₃NH₂ [D] C₆H₅NH₂ [E] NH₃

06:55

### Video Transcript

Which base in the table is most appropriate for the preparation of a buffer solution with a pH of 9.28?

Before we get started with the question, let’s just make sure we know the names for all the chemicals. A is dimethylamine. B is trimethylamine. C is methylamine. D is aniline, otherwise known as phenylamine. And E is ammonia. The question tells us that these five candidates are bases, meaning they’re substances that react with acids.

A buffer solution resists changes in pH. And we’re being asked to pick the best base out of these five candidates to make a buffer solution with a pH of 9.28. We’ve been given the ionization reaction of the base, where the base is reacting with water, protonating itself and producing hydroxide ions.

The equilibrium constant for this process has a special name. It’s the base dissociation constant. The higher the value of K b, the stronger the base. We can use the base dissociation constants to rank the bases. The strongest base in this set is dimethylamine, followed by methylamine, trimethylamine, ammonia, and then aniline.

The question mentions a pH of 9.28. Now pH is a measure of the concentration of H⁺ ions. In water, water molecules are constantly reacting, producing hydrogen ions and hydroxide ions. H₃O⁺ is sometimes written as H⁺ for convenience. The equilibrium constant for this process is given the symbol K w. At the same temperature that our K bs are given, 25 degrees, the value of K w is about one times 10 to the minus 14. And it’s equal to the concentration of H⁺ multiplied by the concentration of OH⁻.

When we introduce a weak base into water, we shift this equilibrium. So we can use the concentration of OH⁻ from these weak bases to work out the concentration of H⁺. And then we can get back to the pH.

The other way of doing it is to turn our base dissociation constant on its head and get the equivalent acid dissociation constant of the conjugate acid. For dissociation constants measured in water, the acid dissociation constant multiplied by the base dissociation constant is equal to K w. For instance, with ammonia, we can figure out the acid dissociation constant of the ammonium ion from K w and the K b of ammonia.

So that’s what I’m going to do. I’m gonna figure out the acid dissociation constant for the conjugate acid of each weak base. The K a for dimethylammonium is 1.73429 times 10 to the minus 11. The K a for trimethylammonium [trimethylamine] is 1.62427 times 10 to the minus 10, and so on. This gives us the highest value of K a for the conjugate acid of aniline, which is what we would expect since aniline is the weakest base. So why have we come all this way?

Well, I’m gonna do a little recap about buffers. Buffers are solutions where we have a weak acid or base and its conjugate. For instance, we might have ammonia and its conjugate acid in the form of ammonium chloride. Buffers are at their best when there’s a one-to-one ratio between the weak acid or base and its conjugate. For our example, ammonia will react with any acid, helping to neutralize it and keeping the pH from fluctuating downwards. Meanwhile, ammonium chloride will react with any base, stopping the pH from rising.

The Henderson–Hasselbalch equation tells us the relationship between the pH of a buffer, the pK a of the weak acid, and the concentrations of the weak acid and its conjugate base. In the case where the concentration of the two components are equal, then the logarithm of their ratio is equal to zero. So in the ideal case where we have the best buffer possible, where we have a 50–50 mix that are weak acid and our conjugate base, the pH of the buffer will be the same as the pK a of the weak acid.

So to find the best base for our buffer, we need to calculate the pK a of that conjugate acid. The pK a is simply the negative logarithm to the base 10 of the K a. The pK a of dimethylammonium is about 10.8. The pK a of trimethylammonium is about 9.8, and so on.

Our job now is to look at all the pK a and see which one is closest to our target pH. The conjugate acid of aniline clearly has a pK a that’s way too low. Methylammonium and dimethylammonium have pK a that are too high. And out of trimethylammonium and ammonium, ammonium has a pK a closest to 9.28.

We can use the Henderson–Hasselbalch equation to figure out concentrations for ammonia and its conjugate acid to produce a buffer with a pH of 9.28. Here I’ve shown one of the many possible combinations. We have 0.1-molar ammonia and 0.092-molar ammonium chloride. So we’ve demonstrated that, out of the five weak bases given, the one most appropriate for the preparation of a buffer solution with a pH of 9.28 is ammonia.

We could have figured it out the other way using the base equivalent of the Henderson–Hasselbalch equation, where we use the pOH and pK b instead of the pH and pK a. Whichever way you do it, you’ll get the same results. And you’ll arrive at the most appropriate base being ammonia.