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.
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
3 answers
1. What is 5/60?
2. How did you get 10.3 for b/a?
2. How did you get 10.3 for b/a?
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!
Thanks!
2 answers