What's the problem. Just plug and chug.
pH = pKa + log (base)/(acid)
use the Hendeson-Hasselbalch equatio. To calculator the pH of a solution at is
A) 0.15 M on C5H5N and 0.10M in C5H5NHCl.
Kb:1.7x10^-9
B) 15.0 g of HF and 25.0g of NaFin 125 ml of a solution.
Ka 3.5x10^-4
3 answers
I guess I am still stuck on problem a. I know I have to calculate Ka from Kb in order to calculate pH. When I do that, I get a pH over 14 which cannot be. What am I doing wrong?
Here is what I would do. Convert Kb to pKb. That is pKb = -logKb = -log 1.7E-9 = 8.77, then
pKa + pKb = pKw. You know pKw and pKb, solve for pKa.
pKa + pKb = pKw. You know pKw and pKb, solve for pKa.