# analytical chemistry

0.100 M NaOH (in the burette) is titrated with 25.0 mL 0.120 M CH3COOH (pKa = 4.75) in a conical flask. Determine the pH of the solution in the flask after 10 mL of the NaOH has been added.

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3. 39
1. millimols CH3COOH = 25.0 x 0.12 = 3.0
millimols NaOH = 10 x 0.1 = 1
..CH3COOH + NaOH ==> CH3COONa + H2O
I...3.0......0........0..........0
C...-1.0....-1.0......+1.0.......
E....2.0......0.......1.0

Substitute the E line into the Henderson-Hasselbalch equation and solve for pH.

pH = pKa + log (base)/(acid)
pH = 4.75 + log (1/35)/(2/35)

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posted by DrBob222

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