TBR chapter on Buffers says max ratio between conj. base and acid can be 10:1

[clothcut]

However, I just encountered a problem on AAMC#4 in which I had to calculate the initial concentrations of a buffer solution set at pH 8.7 using the henderson-hasselbach equation and the ratio/correct answer was 100:1 for ratio of conj. base to acid in solution. How is this possible??? doesnt this mean it is NOT a buffer??

the q says: the pka for the disassociation of h2po4- to hpo4 is 6.7. what is the initial ratio of hpo4- to h2po4- in the buffer solution of experiment 1 (where ph was buffered at 8.7)

thanks

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[milski]

There is not anything magical happening at 1:10 ratio. You get the best buffering characteristics at 1:1 and as the ration increases, the buffering capabilities decrease. Where exactly you want to stop calling the solution a buffer is somewhat arbitrary choice.

 

[BerkReviewTeach]

However, I just encountered a problem on AAMC#4 in which I had to calculate the initial concentrations of a buffer solution set at pH 8.7 using the henderson-hasselbach equation and the ratio/correct answer was 100:1 for ratio of conj. base to acid in solution. How is this possible??? doesnt this mean it is NOT a buffer??

the q says: the pka for the disassociation of h2po4- to hpo4 is 6.7. what is the initial ratio of hpo4- to h2po4- in the buffer solution of experiment 1 (where ph was buffered at 8.7)

thanks

I think the issue with this question is that the HPO₄^2- with H₂PO₄^- mixture is not acting as the buffer here. What they are saying is that if you have a solution that is buffered at 8.7 (which is their way of saying it is held relatively constant at pH = 8.7), then if you were to add a mixture of HPO₄^2- with H₂PO₄^-, it would adjust its equilibrium to be 100 parts HPO₄^2- and 1 part H₂PO₄^-. You can use the Henderson-Hasselbalch equation to get that ratio, but technically speak, those two species do not form a buffer.

A buffer must have a weak acid and conjugate weak base in a ratio within the range of 1:10 to 10:1, otherwise you don't have enough of one of the two species to offset any pH change when adding a strong acid or strong base.

In other words, if you were to mix 100 parts HPO₄^2- with 1 part H₂PO₄^- in water, the solution would have a pH = 8.7, but it would not be a buffer because if you added OH^- the pH would rise drastically (due to the lack of weak acid present to neutralize it).

 

[adrakdavra]

which passage in aamc 4?

 

[preMedDonut]

How I was thinking of this is like this. Since the pka is 6.7 and the pH is 8.7 after buffered we know from the Henderson hassle bach equation that we need to add 2 to 6.7.

The log of 100/1 is 2.

 

[Tarasaurusrex]

Instead of worrying about the buffer ratio definition, jiust solve the problem first.

Write the Henderson-Hasselbach Equation:

pH = pKa - ln(A/HA) where A is the acid and HA is the conjugate base

pH = 8.7 and pKa = 6.7

8.7 - 6.7 = -ln(A/HA)

2 = -ln(A/HA)

-2 = ln(A/HA)

10^-2 = 10^ln(A/HA)

1/100 = A/HA, so ratio of conjugate base to acid is 100:1

 

[letsgetstarted1234]

I think the issue with this question is that the HPO₄^2- with H₂PO₄^- mixture is not acting as the buffer here. What they are saying is that if you have a solution that is buffered at 8.7 (which is their way of saying it is held relatively constant at pH = 8.7), then if you were to add a mixture of HPO₄^2- with H₂PO₄^-, it would adjust its equilibrium to be 100 parts HPO₄^2- and 1 part H₂PO₄^-. You can use the Henderson-Hasselbalch equation to get that ratio, but technically speak, those two species do not form a buffer.

A buffer must have a weak acid and conjugate weak base in a ratio within the range of 1:10 to 10:1, otherwise you don't have enough of one of the two species to offset any pH change when adding a strong acid or strong base.

In other words, if you were to mix 100 parts HPO₄^2- with 1 part H₂PO₄^- in water, the solution would have a pH = 8.7, but it would not be a buffer because if you added OH^- the pH would rise drastically (due to the lack of weak acid present to neutralize it).

bumping this, so can you us henderson hasslebach anytime? i thought you could only use it when you are dealing with a buffer?

 

[NextStepTutor_1]

Hi @letsgetstarted1234 -

You can use the HH equation whenever you have information about two of the three quantities it describes (pH, pKa, and ratio of the concentration of [A-] and [HA]) and want to figure out something about the remaining variable. It tends to be used most often in the context of buffers because that's where it's interesting and useful. For instance, if you took a weak acid with something like a pKa of 7, and you place it in a solution that is super-acidic (with a pH of 1, for instance), that acid will overwhelmingly exist in its protonated form, which is pretty much common sense. Correspondingly, if you put that weak acid in an extremely basic solution (say w/ a pH of 13), virtually all of it will be deprotonated. That's relatively straightforward and not very interesting chemically, but you could use the HH equation to answer a question like "acid XYZ has a pKa of 7, is placed in an unknown solution, and its conjugate base is found to be present in a 10,000:1 ratio with its conjugate acid -- what's the pH of that solution?". That wouldn't be a very interesting question, because you could easily guess that the solution would have to be very basic for this to be the case. The math is less obvious in the context of a buffer, and buffers are also tested more often b/c of their physiological importance, which is why you usually see the HH equation in that context.

Hope this helps!