Berkeley Review pH of a buffer, working with equivalents...

[MCAT guy]

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So here is a question from buffers and titrations general chem #27:

The BEST choice for a pH = 8.5 buffer would be which of the following?

A.
B.
C.
D. H3CCOC6H4OH [read: pKa 8.4] with less than one full equivalent of NaOH

Here is their explanation,

The best choice for a buffer of pH = 8.5 is a weak acid with a pKa close to 8.5 mixed in solution with its conj base. The acid has a pKa of 8.4, which is closest to the 8.5 value. To make the buffer equal to 8.5, there should be a slight excess of the conj base relative to the acid. The log of the ratio of conj base to weak acid should be 0.1, a positive number, so the ratio of conj base to weak acid must be greater than 1

I actually understand their explanation, but I am struggling with the wording of the answer. I would think that you need to add 1/2 an equivalent of the NaOH, by adding the 1/2 you would get the acid and conj base concentrations equivalent and the pKa = pH... I know 1/2 is less than one, but the choice seems odd. Then they go off explaining that you want slightly basic conditions to hit 8.5, but that would mean a little more than 1/2 an equivalent of NaOH correct?

I guess the word equivalent has thrown me off a bit, what is the definition? Just an equal amount? Like 1 equivalent of base added to 50 mL of 1M acid would be 50mL of 1M base?

 

[MCAT guy]

Wait, is 1 equivalent of 100 mL 1M strong acid, 200 mL of 0.5M strong base?

 

[RogueUnicorn]

yes provided both are singly active species (e.g. monoprotic acid)

 

[Phantastic]

Wait, is 1 equivalent of 100 mL 1M strong acid, 200 mL of 0.5M strong base?

Yes, equivalents are in moles, which are the units of the dilution equation, which can be used to find equivalents.

M1V1 = M2V2

0.1*1 = 0.2*0.5 = 0.1
moles of acid = moles of base = 0.1 moles

edit: sorry had the page open from a while ago and got sidetracked