Question

Prepare 100ml of an acetic acid and sodium acetate buffer w/ a pH of 4.50.
HC2H3O2 (aq) = 0.53M
NaC2H3O2 (s) = 136.08g/mol
Ka HC2H3O2 = 1.8 * 10^-5

Answer 1

So far, this is what I've figured out:
\[4.50=4.7+\log \frac{ A ^{-} }{ 0.53M }\]
\[-0.20=\log \frac{ A ^{-} }{ 0.53M }\]
\[0.33=A ^{-}\]

So I need 0.33 moles of the sodium acetate and 0.53 of the acetate to make a solution with a pH of 4.50. Now I need to change these values so I can make 100ml of the buffer. How can I do that?

Answer 2

Does density have anything to do with this?

Answer 3

@AloneS @sweetburger

Answer 4

@ganeshie8

Answer 5

@Photon336

Answer 6

Anyone?