Chemistry Moles of a conjugate base in a buffer solution

AI Thread Summary
To create a buffer solution with a pH of 5.70 using 0.0200 moles of acetic acid, the Henderson-Hasselbalch equation is applied. The pKa of acetic acid is calculated to be 4.74. The discussion emphasizes that the specific volume of the solution is not necessary for determining the moles of sodium acetate needed, as the ratio of concentrations is equivalent to the ratio of moles. Participants clarify that the volume cancels out in the calculations, simplifying the process. The importance of understanding the underlying concepts rather than relying solely on provided data is highlighted.
alr1014
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Homework Statement


You need to produce a buffer solution that has a pH of 5.70. You already have a solution that contains 0.0200 moles of acetic acid. Using the Henderson-Hasselbalch equation calculate the moles of sodium acetate needed to create a buffer with the desired pH? The Ka of acetic acid is 1.8 × 10^-5.

Homework Equations


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

The Attempt at a Solution


pKa= -log (1.8 × 10^-5) = 4.74
5.70 = 4.74 + log (base/acid)
Unfortunately, I have been staring at this problem for about an hour and am unsure how to find the concentration of the acid to discover the consent ration of the base. I am not sure how to proceed since the problem does not give a volume to find the molarity.
Any help with this would be greatly appreciated!
 
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Hint: you don't need the acid concentration.

C_A = \frac {n_A} {V}
C_B = \frac {n_B} {V}

\frac {C_A}{C_B} = ...
 
I appreciate your responce, any responce really. Although, your equation set seems to need a volume as well, and the problem I am to work out does not include one. That is a big area I am being messed up. As well as the problem states to use that particular equation to find the moles of base needed.
 
alr1014 said:
Although, your equation set seems to need a volume as well

Plug concentrations into ratio and see what happens. V/V = ?
 
Borek said:
Hint: you don't need the acid concentration.

C_A = \frac {n_A} {V}
C_B = \frac {n_B} {V}

\frac {C_A}{C_B} = ...
Since the problem called for using the henderson-hasselbalch equation, and that entails using the concentration of the acid, I used the volume used in the lab for the titration (25 mL). Not sure if that is right... but I got an answer. I will check with a few of my class mates before lab in the morning. Thank you for your help.
 
You don't need any specific volume.

V/V = 1 - volume cancels out. Ratio of concentrations (if these are concentrations in the same solution) is identical to the ratio of numbers of moles.
 
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Borek said:
You don't any specific volume.

V/V = 1 - volume cancels out. Ratio of concentrations (if these are concentrations in the same solution) is identical to the ratio of numbers of moles.
That very simple answer was probably the best answer you could have given me! Tha k you so very much for having the patience for dealing with slightly dense people like me!
 
alr1014 said:
That very simple answer was probably the best answer you could have given me!

I hoped you will see it by yourself. You will learn much more when you put an effort and find such things by yourself, than when they are presented on a silver plate.
 
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