Determining pH using Henderson-Hasselbach equation?

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In summary, 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, is 9.4.
  • #1
Vicinity24
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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
 
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  • #2
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.
 
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  • #3
epenguin said:
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.
 
  • #4
From the data given 9.10 should be a correct answer, if it is not accepted I suspect a mistake.

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

1 ± 0.05 ≈ 1
 
  • #5
Borek said:
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.
 
  • #6
Is it possible that I have to multiple the amount of base by 3 because we're given this picture:
kaInpJ9.jpg
 
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  • #7
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?
 
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  • #8
epenguin said:
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.
 
  • #9
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 pKa the only way I can see for the pH not to be exactly that would be to take into account the lower pKa which is about 7. Were you given a pK and were you given only one?
 
  • #10
epenguin said:
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 pKa the only way I can see for the pH not to be exactly that would be to take into account the lower pKa which is about 7. Were you given a pK and were you given only one?

Only given the one pKa I am afraid.
 
  • #11
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.
 
  • #12
epenguin said:
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
 
  • #13
epenguin said:
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.
 

Related to Determining pH using Henderson-Hasselbach equation?

1. What is the Henderson-Hasselbach equation?

The Henderson-Hasselbach equation is a formula used to calculate the pH of a solution based on the concentration of its acid and conjugate base. It is written as pH = pKa + log([conjugate base]/[acid]).

2. How is the Henderson-Hasselbach equation derived?

The Henderson-Hasselbach equation is derived from the dissociation constant (Ka) of an acid and its conjugate base. It is based on the law of mass action, which states that the ratio of products to reactants in a chemical reaction is constant at equilibrium.

3. What is the significance of the pKa value in the Henderson-Hasselbach equation?

The pKa value represents the negative logarithm of the dissociation constant (Ka) of an acid. It is a measure of the strength of an acid, with lower pKa values indicating a stronger acid and higher pKa values indicating a weaker acid.

4. Can the Henderson-Hasselbach equation be used for all acids and bases?

No, the Henderson-Hasselbach equation is only applicable for weak acids and their conjugate bases. Strong acids and bases, which completely dissociate in solution, do not have a dissociation constant and therefore cannot be used in this equation.

5. How accurate is the Henderson-Hasselbach equation in determining pH?

The Henderson-Hasselbach equation is most accurate for solutions with a pH close to the pKa value of the acid. As the pH moves further away from the pKa, the accuracy decreases. Other factors, such as temperature and ionic strength, can also affect the accuracy of the Henderson-Hasselbach equation.

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