# CHEMISTRY

Calculate the amount (in mL) of 1.520M NaOH that is required to add the following acetic acid solution to prepare a buffer with the corresponding pH:

30.00mL of a 5.00% (w/v%) acetic acid; the resulting acetate buffer has a pH of 5.75.

pKa of acetic acid = 4.74

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3. 20
1. I would do this.
5% w/v means 5g HAc/100 mL solution which is (5/60) mols/100 and that is 5/60/0.1L or 0.0833 M.
Then millimols HAc is 30 x 0.0833 = 25.
So acid + base = 25
pH = pKa + log(base)/(acid)
5.75 = 4.74 + log b/a
b/a = about 10.3 but you need a better answer than that so go through it yourself.That's equation 1.
equation 2 is a + b = 25
Solve the two equation simultaneously for a and b. I get something like 2 for acid and 23 for base.
Then knowing that M = millimols/mL.
1.52 = 23/mL; solve for mL = about 15 mL of the 1.52 M NaOH. Of course that is about 10 for the acid that is left.
To check all of this I would plug the acid you find and the base you find into the HH equation and see if the final pH is 5.75.

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posted by DrBob222
2. 1. What is 5/60?
2. How did you get 10.3 for b/a?

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posted by Jeff
3. Hi DrBob222, can you explain what you mean by "Solve the two equation simultaneously for a and b. I get something like 2 for acid and 23 for base." I don't know how you got that and what you started with.
Thanks!

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posted by Jeff

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