TBR CHEM-I P.300 (Ex 5.8) Buffer/pH Question

[Rose1Line]

---

Members do not see this ad.

Hi Everyone,
Could someone please show me how the following question (in TBR Chem-I page 300, Example 5.8) is calculated? I am getting the correct answer (B. 8.26) by process of elimination, but not sure how to derive at this answer.

Q: A buffered solution initially has a pH of 8.31. When five drops of 12M HCl are added to a 500-mL beaker filled with this buffered solution, what would be expected for the final pH value?

A. 3.31
B. 8.26
C. 8.31
D. 8.36

Thanks in advance 🙂

 

[Phantastic]

No calculation necessary, just intuition.

Check it out:

Obviously, the answer is not C or D, because the pH should go down...at least by one one-hundredth of a point...

Now, does it go all the way to pH=3 with just five drops? If you've seen a calculation in the past involving a strong acid added to a solution of pure water, you may be aware of how powerful 5 drops can be (it could lower it that far with just 0.02 mL of 12 M HCl)...BUT, since your solution here is at a pH greater than 7, their is already some OH- dominating in the solution, and the initial H+ will merely bind to that to produce water, lowering the pH slightly due to less OH- in solution.

Hope that helps.

 

[IntelInside]

Henderson-Hesselbach equation could be used here if we knew the pKa of the acid or its pKb of the base

pH = -log([H+)
pH + pOH = 14
pOH = -log([OH-]

Then you would have to calculate the new molarity of the added H+ ions via the dilution formula (M1V1=M2V2) (remember we dont know the volume added of those 5 drops), and then adjust the total molarity of H+. Then you have your answer via:

pH = pKa + log ([Base]/[Acid])

or

pH = 14 - (pKb + log ([Acid]/[Base])


Those calculations are way to lengthy for the MCAT I believe. Intuition is your best friend sometimes for the MCAT

 

[Rose1Line]

No calculation necessary, just intuition.

Check it out:

Obviously, the answer is not C or D, because the pH should go down...at least by one one-hundredth of a point...

Now, does it go all the way to pH=3 with just five drops? If you've seen a calculation in the past involving a strong acid added to a solution of pure water, you may be aware of how powerful 5 drops can be (it could lower it that far with just 0.02 mL of 12 M HCl)...BUT, since your solution here is at a pH greater than 7, their is already some OH- dominating in the solution, and the initial H+ will merely bind to that to produce water, lowering the pH slightly due to less OH- in solution.

Hope that helps.

Oh that totally makes sense, thanks.

 

[Rose1Line]

Henderson-Hesselbach equation could be used here if we knew the pKa of the acid or its pKb of the base

pH = -log([H+)
pH + pOH = 14
pOH = -log([OH-]

Then you would have to calculate the new molarity of the added H+ ions via the dilution formula (M1V1=M2V2) (remember we dont know the volume added of those 5 drops), and then adjust the total molarity of H+. Then you have your answer via:

pH = pKa + log ([Base]/[Acid])

or

pH = 14 - (pKb + log ([Acid]/[Base])


Those calculations are way to lengthy for the MCAT I believe. Intuition is your best friend sometimes for the MCAT

Thanks IntelInside for the calculation! I'm glad to hear that this is probably too lengthy for the MCAT b/c I was panicking!

 

[BerkReviewTeach]

Thanks IntelInside for the calculation! I'm glad to hear that this is probably too lengthy for the MCAT b/c I was panicking!

They don't give you the concentrations of the buffer components or the pKa of the weak acid, so you couldn't solve it using math anyway. And the calculation would be lengthy if they did. This is one of those questions that makes most people panick at first. Trusting your intuition rather than going through a process to arrive at an answer is an essential part of getting ready for this exam. Cutting down your time per question and finding the most efficient route to the answer should be one of your study goals. Good luck! And you're almost half way done with G Chem, YEAH!!!