4.80 = pKa + log(x)/(0/150)
Substitute pKa and solve for x.
What CH3COONa concentration is required to prepare a buffer solution with a pH of 4.80 if [CH3COOH] = 0.150 M? Ka of CH3COOH = 1.8 10-5.
4 answers
PH=-log(ka)+log(base/acid)
4.8=-log(1.8*10^-8)+log(x/.15)
4.8=7.7447+log(x/.15)
-2.9447=log(x/.15)
.001135=x/.15
x=1.875*10^-4
4.8=-log(1.8*10^-8)+log(x/.15)
4.8=7.7447+log(x/.15)
-2.9447=log(x/.15)
.001135=x/.15
x=1.875*10^-4
If I use your numbers I come out with 1.7E-4. The difference arises from 0.001135*0.15 = 1.7E-4 and I assume you just punched a wrong button; however, none of that counts because you didn't use the right Ka.
Ka = 1.8E-5 so pKa = 4.74
Ka = 1.8E-5 so pKa = 4.74
sorry read it wrong, doing this now in chem
2 answers