Determining pH using Henderson-Hasselbach equation?

[Vicinity24]

Homework Statement

Determine the pH of the solution (to 2 decimal points) after the addition of 50.0 mL of 2.5 M nitric acid (HNO3) to 500 mL of 0.500 M BIS−TRIS propane (C11H26N2O6), a weak base). Assume that the 5% approximation is valid and that the volumes are additive. The pKa of BIS−TRIS propane−H is 9.10.

Homework Equations

Henderson-Hasselbach equation:
pH = pKa + log(nbase/nacid)
n being amount of moles

The Attempt at a Solution

moles acid= (50.0mL/1000mL)*2.5M= .125 moles of acid

moles base= (500mL/1000mL)*.5M= .250 moles of base

Using henderson-hasselbach equation:

pH = 9.10 + log(.250/.125)

pH = 9.4

It is however, incorrect. I've had to do a set of these and I got those correct but not this one following the same format. I was curious if maybe I need to multiple the base by a certain molar ratio? I could not figure it out given the complicated equations though.

Thanks

 

[epenguin]

Probably you have confused what is meant by base and acid here - it means the moles or molarities of the basic and acidic forms, B and BH⁺ of the buffer. You have added half as many moles of acid as there are of base so the molarities of these two forms are equal.
Their ratio is 1, log(1) = 0, so pH = pK

I do not know what "the 5% approximation" is.

But note in case there are other questions on this substance that bis-tris-propane is dibasic, however the first pK is more than two units below the second, at pH 9.1 less than ½% of it would be diprotonated (BH₂²⁺) and you do not need to worry about it.

 

[Vicinity24]

Probably you have confused what is meant by base and acid here - it means the moles or molarities of the basic and acidic forms, B and BH⁺ of the buffer. You have added half as many moles of acid as there are of base so the molarities of these two forms are equal.
Their ratio is 1, log(1) = 0, so pH = pK

I do not know what "the 5% approximation" is.

But note in case there are other questions on this substance that bis-tris-propane is dibasic, however the first pK is more than two units below the second, at pH 9.1 less than ½% of it would be diprotonated (BH²⁺) and you do not need to worry about it.

Using a ratio of one, it's still incorrect though.

You don't need to know the 5% approximation for this question as there were a bunch of others where I didn't use it. Also the last thing you said goes beyond what I know right now as this is an introductory chem course in university.

 

[Borek]

From the data given 9.10 should be a correct answer, if it is not accepted I suspect a mistake.

I do not know what "the 5% approximation" is.

1 ± 0.05 ≈ 1

 

[Vicinity24]

From the data given 9.10 should be a correct answer, if it is not accepted I suspect a mistake.
1 ± 0.05 ≈ 1

Strange, well thanks for the help.

 

[Vicinity24]

Is it possible that I have to multiple the amount of base by 3 because we're given this picture:

 

[epenguin]

No, there is an N atom there which can be either protonated or not, according to the pH, that's all.
There is another such atom but it can be ignored for present purposes for reasons already given,

Assuming that, two of us agree that the pH should = pK = 9.1 .

If you are told this is wrong have you been told any other answer?

 

[Vicinity24]

No, there is an N atom there which can be either protonated or not, according to the pH, that's all.
There is another such atom but it can be ignored for present purposes for reasons already given,

Assuming that, two of us agree that the pH should = pK = 9 1 .

If you are told this is wrong have you been told any other answer?

I've got one try left and it doesn't show me the correct answer until after it's due which is a week.

 

[epenguin]

If we have at the same time this 5% rule and we have to determine pH to 2 decimal places it's possible we can't do it.

If you were given a pK_a the only way I can see for the pH not to be exactly that would be to take into account the lower pK_a which is about 7. Were you given a pK and were you given only one?

 

[Vicinity24]

If we have at the same time this 5% rule and we have to determine pH to 2 decimal places it's possible we can't do it.

If you were given a pK_a the only way I can see for the pH not to be exactly that would be to take into account the lower pK_a which is about 7. Were you given a pK and were you given only one?

Only given the one pKa I am afraid.

 

[epenguin]

Well I calculate that this second protonation doesn't change even the second decimal anyway, so all I can suggest is if you answered 9.1 answer 9.10, but if you already answered that and it was held 'wrong', answer 9.1

Will you be given an explanation of why your answers are wrong? I so please come back and tell us.

 

[Vicinity24]

Well I calculate that this second protonation doesn't change even the second decimal anyway, so all I can suggest is if you answered 9.1 answer 9.10, but if you already answered that and it was held 'wrong', answer 9.1

Will you be given an explanation of why your answers are wrong? I so please come back and tell us.

My answer was 9.10, it doesn't mark it without two decimal places anyways. I'll get the answer with no explanation. I will report back. I

 

[Vicinity24]

Well I calculate that this second protonation doesn't change even the second decimal anyway, so all I can suggest is if you answered 9.1 answer 9.10, but if you already answered that and it was held 'wrong', answer 9.1

Will you be given an explanation of why your answers are wrong? I so please come back and tell us.

Okay so turns out since it's a strong acid you need to do:

pH=pKa+log((moles base - moles acid)/(moles acid))

But you get zero leading to 9.10 for my case. Not sure why it's like this it didn't show the answer either, probably because it's the last assignment of the semester. But anyways, people reading this from the following years stuck on this problem, just use the equation above and you'll be fine.