Need help with question. Calculate the pH of a buffer prepared by adding 100mL of 0.0500M NaOH to 100mL of 0.175M CH3COOH.

Question

so far I have calculated mols of
CH3COOH---->H+ + CH3OO-

CH3COOH=0.1L*0.175mol/1L=0.0175mol
OH-=0.1L*0.0500/1L=0.00500mol
Not sure were to go from here, or if the above is right

Answer 1

Millimoles is a little easier to work with in these problems; it keeps all those leading zeros out of the way.
100 mL x 0.175M CH3COOH = 17.5 mmoles.
100 mL x 0.05M NaOH = 5 mmoles.

.......NaOH + CH3COOH ==> CH3COONa + H2O
init...5.0....17.5.........0..........0
change.-5.0...-5.0.......+5.0
equil...0....12.5.........5.0

So you have CH3COOH and CH3COONa (a weak acid and its salt which gives you a buffer). Use the Henderson-Hasselbalch equation to solve for pH. I get about 4.3 or so but you need to do it more accurately.
pH = pKa + log[(acetate)/(acid)]

Answer 2

How do i solve if i am not aloud to use henderson-hasselbach?