pH = pKa + log( [A-]/[AH])
Plug in the values for pKa and plug in 3.5 for [A-]/[AH] and solve.
pH = 4.74 + log(3.5)
For the second part of the question, I am not sure if it is asking for the concentration of acetic acid to prepare a buffer solution with a pH of 4. But if it is then here is the setup:
pH = pKa + log( [A-]/[AH])
The answer for [A-]/[AH] multiplied by 0.6 should give you the number of moles of acetate, and subtracting the number of moles from 0.6 should give you the moles of acetic acid needed.
Again, I'm not sure if it is asking for concentration of acetic acid. Maybe Dr. Bob222 will check this and correct it if needed.
Remember, the number of moles/liters= molarity
Something may not have been clear the second line from the bottom should say
The answer for [A-]/[AH] multiplied by 0.6 should give you the number of moles of acetate, and THE NUMBER OF MOLES OF ACETATE CALCULATED SUBTRACTED from 0.6 should give you the moles of acetic acid needed.
I showed you how to do both of these yesterday.
I do apologize DrBob222 I didn't see your post from yesterday. However, thank you for help and thank you devron for your help as well. The extra credit only says how would prepare a .6m acetate buffer with a pH = 4.0. So I too, wasn't sure what the question was asking. However, thank you all for your outstanding explanation. It really helped me :)
I just looked at Dr. Bob222's post for when you initially posted this question and the setup that I gave you for part 1 differs from the setup that he gave you for part 2. I will let him resolve that for you since he has more experience in G-chem then I do, and he is probably correct anyway.
I mean are setups differ for part 1
One confusing part in biochemistry is when they talk about buffers. When I was a student (a looooong time ago) the problems always stated, for example, 0.1M in acetic acid and 0.05 M in sodium acetate so there was never a problem in knowing the concentrations. When I first started helping here and got into the buffer systems I found the statement for example, "the 0.5M buffer......." After reading hundreds of articles and word out Google I finally figures out that when the problem states "0.5M acetate buffer" it really means the total acetate is 0.5M; i.e., acetate + acetic acid = 0.5M. I called a biochemistry friend of mine and confirmed that. I should have called him first. How do you separate the total of 0.5 into the acid part and the base part? Two equations.
(base) + (acid) = 0.5M
The second equation you use is the HH equation.
From pH = pKa + log (base)/(acid) you can solve for the ratio base/acid. That gives you two equations in two unknowns which allows you to calculate the base concn and the acid concn.
You are correct and that is was I conveyed in the setup for the second part as well. Looking back at your original post, I thought you let
CH3COOH = 3x and CH3COO^- = x, but you have CH3COOH = x and CH3COO^- = 3x
But looking at it again, I see that are setups are the same, just different ways of explaining it.
Usually I see these types of problems written as, "How would you prepare a 0.6 M buffer with a pH = 4.0"? The word acetate behind 0.6M, initially led me to believe that the concentration of acetate was 0.6M and that they wanted to know hoe much acetic acid to add, but when I was done with 90% of the setup, I realized that I was correct, but wanted you to double check to make sure; I don't want anyone blaming me for a professor or teacher marking a problem wrong on their homework.
Isn't the ratio
ignore isn't the ratio; I forgot to delete that when I pasted your original post.
organic chem - . Use the Henderson-Hasselbach equation: pH = pKa + log [A-]/[AH...
Science - Given: Concentration of acetic acid 0.1M and 30mL used. Concentration ...
Chemistry - A buffer containing 1.2169 M of acid, HA, and 0.1431 M of its ...
CHEM REPOST...PLEASE CHEK! - Posted by Kat on Monday, March 2, 2009 at 11:21am. ...
Chemistry - If a researcher needs 0.50 L of 3.0 M nitrite buffer at pH 3.00, ...
Biochemistry - A person during a 24 hour period produced 1.40 litres of urine ...
CHEMISTRY HELP PLEASE - when I use henderson-hasselbach Ka of HOP4^2- = 4.8 x10...
PKA VALUES - I understand that the pH at the half equivalence point gives is the...
Chemistry - Hello, I have a question that reads, "What is the pH of a 0.01M (...
chemistry - hello, I can't seem to solve this question. i tried to look up ...