If a buffer solution is 0.100 M in a weak acid ( 𝐾a=4.9×10−5) and 0.440 M in its conjugate base, what is the pH?

Answer 1

I figured this one out for myself haha... directions as follows:

1. -log(Ka)=pKa...-log(4.9x10-5)=4.3098...
2. pH=pKa+log(cA/cHA)
3. pH=4.3098...+log(.440/.100)
ans=4.9532... or 5.0

Answer 2

I didn't do the math but your process is on the mark. Good work.