[xc630]

Would appreciate some guidance with the following problem:

What is the pH of a solution that is 0.50M in sodium acetate and 0.75 M in acetic acid? ka for acetic acid is 1.85 x 10^-5

I did ka= x^2/ (0.75-x)
I assumed 0.75-x=0.75
So 1.85 x 10^-5 =x^x/ 0.75
x=.00372
pH=-log(0.00372)=2.43

Is this correct? I wasn't sure where the 0.50M in sodium acetate comes into play. Is this extra unneeded info? Thanks!

[enahs]

Sodium acetate is the conjugate base of Acetic Acid.

You have a weak acid and its conjugate base, ring a bell?

[xc630]

Not exactly. would it be ka= x (0.5)/ (0.75)?

[Borek]

Not exactly.

Buffer?

would it be ka= x (0.5)/ (0.75)?

That should work OK.

[mattc82]

I worked it and got the following,

1.85x10^-5=(x)(.5)/(.75)

2.78x10^-5=x

pH=-log(2.78x10^-5)

pH=4.56

[Borek]

That's OK, but it won't hurt if you will learn about a Henderson-Hasselbalch equation. You already almost derived it.