Chemistry CR page 179 question 10

Lionroar
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Chemistry CR page 179 question 10

Postby Lionroar » Wed Oct 18, 2017 9:29 am

What is the difference between the equivalence point and the pKa? The answer in the back says that at the pKa, the acid-base ratio is 1. But on page 173, the equivalence point was defined as the point at which equivalent amounts of acid and base have reacted. Isn't that the same thing?
NS_Tutor_Andrew
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Re: Chemistry CR page 179 question 10

Postby NS_Tutor_Andrew » Thu Oct 19, 2017 12:55 pm

Hi Lionroar,

Excellent question! The phrasing here can be a little tricky, and context is important. The pKa is the point at which the amount of an acid equals the amount of its conjugate base, whereas the equivalence point in a titration occurs where the amount of acid/base in the titrant equals the amount of the unknown acid/base. To make this more concrete, imagine that we take a solution of HF, a weak acid, and titrate it with OH-, a strong base. At the pKa, there is an equal amount of HF and F-, and at the equivalence point, the amount of OH added is equal to the amount of HF that was originally present in the solution.

Hope this clarifies things!
Andrew D.
Content Manager, Next Step Test Prep.
Lionroar
Posts: 21
Joined: Thu Sep 28, 2017 3:02 pm

Re: Chemistry CR page 179 question 10

Postby Lionroar » Fri Oct 20, 2017 10:45 am

So, at the end of the day, a solution at the equivalence point and a solution at its pKa will both have a pH of 7, but for different reasons? The former because the added strong base has neutralized the weak acid, while in the latter because the weak acid is halfway dissociated. Thanks.
NS_Tutor_Andrew
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Re: Chemistry CR page 179 question 10

Postby NS_Tutor_Andrew » Fri Oct 20, 2017 2:28 pm

Hi Lionroar,

I'm really glad you asked this question because it provides the opportunity to clear up some important misconceptions!

In reality, a pH of 7 occurs only under some very specific conditions -- for pure water at 25°C and at the equivalence point of a titration of a strong acid w/ a strong base (or vice versa) -- that is, not for all titrations and not when pH = pKa.

Let's start with titrations. If we titrate a strong acid with a strong base, at the equivalence point they will be perfectly neutralized and form a salt that doesn't engage in acid-base chemistry. The pH is 7, life is simple, and we move on. (A classic example is the neutralization reaction of HCl + NaOH --> NaCl + H2O). However, if we titrate a weak acid with a strong base, the pH will be >7 at the equivalence point (i.e., the solution will be basic). The reason for this is that the conjugate base of a weak acid is itself a pretty decent base, so it will deprotonate some H2O, pushing the pH up. Similarly, if we titrate a weak base with a strong acid, the pH at the equivalence point will be <7 (i.e., the solution will be acidic) because the conjugate acid of the weak base is strong enough to protonate some water. The Khan academy article on this is pretty solid and goes into some more detail.

pKa is a whole different story, though. Let's step back and review how the Henderson-Hasselbach equation is derived. This may seem tedious, but it will help you understand the difference more clearly.

(1) We want to measure the strength of an acid. To do so, let's use Ka, which is just a special form of the equilibrium constant, and can be used as an indicator of how much the forward direction of a reaction is favored compared to the reverse reaction. So, for the acid reaction HA --> H+ + A-, we can write Ka = [H+][A-]/[HA]. High values of Ka mean that the forward reaction is favored and that the acid is strong.

(2) Now, Ka will have a huge range of values, and as is often the case for such situations, a logarithmic scale is more convenient. We'll use the "p" notation to define pKa as -logKa, or -log[H+][A-]/[HA]. Now a low pKa means a strong acid.

(3) We can do some mathematical jujitsu here using rules about how to manipulate logarithms: log(xy) = log(x) + log(y). This isn't explicitly tested on the MCAT, so don't sweat it if this seems strange, but this allows us to say pKa = -log[H+] - log[A-]/[HA].

(4) Ooh! Here we notice that -log[H+] is defined as pH. We can now simplify this to pKa = pH - log[A-]/[HA], and then do some algebra to rearrange this to the usual form of the equation, pH = pKa + log[A-]/[HA].

From a purely mathematical standpoint, if [A-] = [HA], then [A-]/[HA] = 1 and log(1) = 0, so pH = pKa. Remember that strong acids will have low pKa values and weak acids will have high pKa values. To return to your question, it's not quite right to say "a solution at its pKa." The pKa is just a measure of how much an acid likes to dissociate, not a point that a solution can be "at". However, a neat fact is that when the concentrations of the acid and its conjugate base are equal, the resulting pH will equal the pKa.

Another way of looking at this is to remember that pH = -log[H+] and to note that in our derivation of the HH equation, what we essentially did was just to pull that term out of the Ka expression, which is why pH and pKa are two different things entirely.

Hope this is helpful! This is sort of a tricky point but it's one worth getting good at, both b/c it can be tested directly and b/c it can be incorporated into passages or as first steps in multistep questions.
Andrew D.
Content Manager, Next Step Test Prep.

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