Posted by Alania on Wednesday, November 3, 2010 at 4:59pm.
The secret to this type problem is to know where you are on the titration curve.
C6H5COOH + KOH ==> C6H5COOK + H2O
For (a),
C6H5COOH + H2O ==> C6H5COO^- + H3O^+
Set up an ICE chart, substitute into Ka expression, and solve for H3O^+, then convert to pH.
For the others.
b,c,d. Calculate moles C6H5COOH initially. Calculate moles KOH added. Subtract moles KOH from moles benzoic acid to arrive at the excess benzoic acid, then substitute into either (1)Ka expression or (2) Henderson-Hasselbalch equation.
e. You find the stoichiometry point by mLa x Ma = mLb x Mb
a = acid; b = base.
f. You are at the equivalence point. The pH is determined by the hydrolysis of the salt, C6H5COOK (C6H5COO^- acting as a base).
C6H5COO^- + HOH ==> C6H5COOH + OH^-
Kb = (Kw/Ka) = (C6H5COOH)(OH^-)/(C6H5COO^-)
Post all of your work if you get stuck.
Set up ICE chart and solve for OH^- and convert to pH.
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