Chemistry
posted by Bob on .
40.0 ml of an acetic acid of unknown concentration is titrated with 0.100 M NaOH. After 20.0 mL of the base solution has been added, the pH in the titration flask is 5.10. What was the concentration of the original acetic acid solution? (Ka(CH3COOH) = 1.8 × 10−5)

let X = M CH3COOH.
(CH3COOH) = (40mL*XM/60 mL)
(NaOH) added = (20 mL x 0.1M/60mL)
........CH3COOH + NaOH ==> CH3COO^ H2O
I.......(40X/60)...0........0..........0
add...............2/60...............
change...2/60... 2/60.....2/60.....2/60
equilib (40X2/60)..0........2/60....2/60
Substitute into the HendersonHasselbalch equation and solve for X.
5.10 = 4.74 + log(2/(40x2) and solve for X 
dkls,45