posted by Shell .
A 30.00 mL volume of a weak acid, HA, (Ka = 3.8 x 10^-6) is titrated with 39.00 mL of 0.0958 M NaOH to the equivalence point.
A.) What is the pH of the acid solution, before any base is added to it?
B.) What is the pH of the acid solution after exactly 19.50 mL of NaOH is added?
HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
A) pure HA.
Set up ICE chart for pure HA, substitute into Ka expression, solve for (H^+) and convert to pH.
moles HA initially = M x L = ??
moles NaOH at 19.5 mL = M x L = ??
Subtract mols HA - moles NaOH. The difference is the amount of HA remaining, moles NaOH is the amount of A^- formed. Convert moles of HA and A^- to molarity, plug into Ka expression, and solve for H^+, then convert to pH. As an alternative, you may plug those concns into the Henderson-Hasselbalch equation and obtain pH directly.