Right. Up to
a + 0.87a = 0.5 but then you went in the wrong direction.
If a + 0.87a = 0.5 then
1.87a = 0.5 and
(acid) - 0.5/1.87 = 0.267 M and
(base) = 0.5-0.267 = 0.233 M.
So M acid = 0.267 = mol/L. You know L and M, solve for mols and convert to grams.
Do the same for M base.
Calculate how you would make 0.5L of a 0.5M (500mM) acetate buffer at a pH of 4.7. Available is sodium acetate and glacial acetic acid.
pH=4.7
pKa of acetic acid=4.76
Not sure if I did this next part right:
4.7=4.76 + log [base]/[acid]
-0.06= log [base]/[acid]
10^-0.06= [base]/[acid]
0.87=[base]/[acid]
0.5= base + acid and
0.87= [base]/[acid]
0.87acid + acid = 0.5
acid + acid = 0.575
I guessed on this part trying to follow along with other examples
10^0.575 = 3.7 ~ 4
acid= 4
0.5= base = 4
base= -3.5
I am really stuck on what to do next...I would appreciate any help!
2 answers
What is the concentration of HOAc in a 0.8M Acetate buffer, pH 2.00?
(pKa for Acetic acid is 4.77)
(pKa for Acetic acid is 4.77)