PH of of a buffer solution? What am I doing wrong?

Question

Alright so this is a question for an assignment I have to do online. When inputting my answer it is telling me it is wrong and I re-did my numbers and I keep getting the same answers, so I am assuming my process may be wrong. Please tell me if you can identify what I am doing wrong.

Question: Calculate the pH of 800.0 mL of solution containing 19.0 g of CH3COOH and 25.0 g of CH3COONa.

So I calculated my concentration values by figuring put the molar mass ands than dividing by 0.8 L. Therefore CH3COOH = [0.395] and CH3COONA = [0.381].

So I assumed that CH3COO- is going to be [0.381] because CH3COONA dissociates completely because it is a weak acid. So my ice table looks like this:

CH3COOH --> CH3COO- + H+
0.395 0.381 O
- x +x + X

So I solved for H+ and got 1.8 times 10 to the negative 5. So I calculated OH value by using KW value of 1E - 14. And than used Ph = -log ( ). I got 9.26 Is it correct?

Nevermind I figures it out. I had to take the - log of my H+ value. :P What a relief!

Answer 1

the molarities are correct
CH3COOH = [0.395]; [CH3COONa]=0.381

you have seen the formula

pH = pKa + log([salt]/[acid] (pka for your acid =4.76)

here pH = pKa+ log 0.381/0.395 = 4.76+log 0.96= 4.76-0.02
pH =4.74