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

Analytical Chemistry 
DrBob222,
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 HendersonHasselbalch 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)] 
Analytical Chemistry 
Anonymous,
How do i solve if i am not aloud to use hendersonhasselbach?