Question

how to calculate pH of a butter composed of 0.12M benzoic acid and 0.20M sodium benzoate

Answer 1

henderson-hasselbalch equation. you will need the Ka of benzoic acid

Answer 2

yes it is..but i dont understand how o use the formula

Answer 3

The Henderson-Hasselbach equation is just the equilibrium constant (Ka) equation for a weak acid, with certain simplifying assumptions that usually hold good in a buffer. Start with the Ka equation:

Ka = [H3O+][A-}/[HA}

in which I've indicated the weak acid by HA and its conjugate base as A-. Hopefully this equation is familiar to you, because it's the one you use to calculate the composition of, and hence pH of, any weak acid solution.

Now let us take the log base 10 of both sides, and remember the properties of logs, e.g. log ab = log a + log b:

- pKa = - pH + log[A-]/[HA]

Note that we've also used the definition of pH = - log [H3O+] and pKa = - log Ka. Solve for the pH:

pH = pKa + log [A-]/[HA].

This is the HH equation, or at least one of its algebraically equivalent forms. Now here comes the assumptions. Ordinarily, to solve an equilibrium problem, you would assign x to one of the unknown concentrations at equilibrium, say [H3O+], and then use the chemical equation to write all the other equilibrium concentrations in terms of their initial concentrations and x, often using an "ICE" table. You end up with an equation in x, usually at least a quadratic, solve it, and then go back and figure out all the equilibrium concentrations.

But we can make some simplifying assumptions here, because this is a buffer. That means we know two things:

(1) HA is a WEAK acid, so the amount of dissociation of HA will be very small.

(2) The initial concentration of HA and A- will be large and roughly equal. (That's how you make a buffer.)

Both of these facts mean that to a good approximation, the equilibrium values of [HA] and [A-] will be hardly different from their initial values. So we can just plug the initial values into the HH equation and solve for pH:

pH = pKa + log (0.20 M / 0.12 M)

Looks like we'll need the pKa of benzoic acid, so we do a little googling and find from Wikipedia that pKa = 4.21. If I were you I'd double check that with my chemistry textbook.

pH = 4.21 + log (0.20/0.12) # notice the units cancel -- a good sign.

pH = 4.21 + log (1.666667) # two sig dig in result of division.

pH = 4.21 + 0.5108... # two sig places in result of log.

pH = 4.72 # two sig places in result of addition.

How about a sanity check? We end up with a pH that is slightly higher (less acidic) than the pKa of the acid, but that makes sense, because we have more benzoate (the conjugate base) than benzoic acid. In fact, we can see from the equation itself that the more benzoate we have, the higher the pH, and contrariwise the more benzoic acid we have, the lower the pH (and more acidic the buffer).

I mention the last point because one of the easiest mistakes to make using the HH equation is to get the sign in front of the log wrong, or which concentration ([HA] or [A-]) goes on top of the fraction wrong. You can not work at remembering this perfectly, as long as you can use your physical intuition to correct the sign or fraction as needed, by remembering that a bigger [HA] (more acid) should bring the pH down from the pKa, and a bigger [A-] (more base) should bring the pH up.