[thegoodaaron]

I'm having a hard time working on the follow problem:

What volumes of 0.43 M HNO2 and 0.57 M NaNO2 must be mixed to prepare 1.00 L of a solution buffered at pH = 3.66?

I'm assuming that this problem should be approached using a Henderson-Hasslebalch setup like this:

pH = pKa + log ([base]/[acid])
3.66 = -log(Ka for HNO2) + log ([base]/[acid])
([base]/[acid]) = 1.82
1.8 x Vol(acid) = Vol(base)
Vol(acid) + Vol(base) = 1L
Vol(acid) + 1.8Vol(acid) = 1L
Vol(acid) = 0.36L
Vol(base) = 0.64L

This, however, is not correct. Where have I gone wrong? Thanks in advance for any comments/suggestions.

[Borek]

This is correct approach. Don't round intermediate results - 1.82 gives me 0.35, not 0.36 as a final answer.

What pKa value have you used?

[thegoodaaron]

I used the pK_a value from my textbook (and therefore it should be correct for the problem), which is equal to 4.0e-4. Perhaps I am simply messing up my rounding/calculations? I am not particularly good with logs. Also, I tried both .35 and .36 and neither were accepted. Thanks!

[Borek]

Sorry, shouldn't post before coffee. I missed one thing. You have omitted concentrations of the original solutions, what you did will work only if both solutions concentrations are identical. The are not.

[thegoodaaron]

How should they be factored in? Multiply .35L by the original [HNO2] and .65 by the original [NaNO2]?

[Borek]

Look at the H-H equation - expression under log. You can cancel volumes and you are left with ratio of numbers of moles. One of your equations should be just sum of volumes, while the second should reflect this number of moles ration.

In other words: just use concentration definition to modify your calculations.

[english]

The ANS key is your friend.  Prevents rounding error.

[thegoodaaron]

Look at the H-H equation - expression under log. You can cancel volumes and you are left with ratio of numbers of moles. One of your equations should be just sum of volumes, while the second should reflect this number of moles ration.

In other words: just use concentration definition to modify your calculations.

I'm sorry, but is it possible to explain this in another way? I am far too easily confused. Thank you!

[Borek]

[base]/[acid] = 1.82 - that's OK.

Vol(acid) + Vol(base) = 1L - that's OK too. But

1.8 x Vol(acid) = Vol(base) - this one is wrong.

In the final solution you need

Vol(buffer) x [acid]

moles of acid and

Vol(buffer) x [base]

moles of base. These amounts will be present in the 0.43 x Vol(acid) and 0.57 x Vol(base) respectively. Try to use this information.

[thegoodaaron]

I'm still confused and answerless.

So we have decided that Vol(acid) + Vol(base) = 1.000L

and that

[base]/[acid] = 1.82 (actually, 1.83 now that I do it again)

But I feel like I am still missing some crucial step in my mind to tie it all together.

You noted that

Vol(buffer) x [acid] = moles of acid and

Vol(buffer) x [base] = moles of base.

However, I'm not sure how to use this information. The volume of the buffer solution is 1.000L. The initial concentrations of acid and base are, respectively, .43M and .57M. Multiplying 1.000L by each of these values leaves us with .43 mol acid acid and .57 mol base. But where does the 1.83 come into play? How is that factored in? Thank you for being patient with me.

[Borek]

The initial concentrations of acid and base are, respectively, .43M and .57M. Multiplying 1.000L by each of these values leaves us with .43 mol acid acid and .57 mol base.

Buffer solution concentrations differ from 0.43 and 0.57!

You have to set up two equations - one for sum of volumes, second for the concentration ratios. Just be precise when you have buffor concentrations (ratio 1.83) and when you have initial concentrations (0.43 and 0.57).

[thegoodaaron]

Thank you endlessly. I appreciate your time and patience. I got it!