Practice Mid2 09 soln corr prob 4

Practice Mid2 09 soln corr prob 4 - log (X / 100 – X) =...

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4 3. (a) (6 marks) You want to make a buffer of pH = 7.0, how much 0.20 M NaOH is needed if you have 250.0 mL solution of 0.40 M acetic acid? Completion reaction in mmol (can also make the table in moles): CH 3 COOH (aq) + OH- (aq) CH 3 COO - (aq) + H 2 O (l) i. 100 X 0 c. -X -X +X comp 100-X 0 X OH - has to be the limiting reagent if you are going to make a buffer. Use Henderson-Hasselbach equation: pH = pKa + log ([CH 3 COO - ]/[CH 3 COOH]) ; Ka = 1.8 × 10 -5 ; pKa = 4.74 7.0 = 4.74 + log ( X /100-X ); don’t need to convert to concentrations as volume in ratio cancels
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Unformatted text preview: log (X / 100 – X) = 2.2 6 X / 100 – X = 10 2.5 6 X / 100 – X = 18 2 ; X = 99.4 5 mmoles of NaOH needed Volume NaOH needed = moles / concentration = 99.4 5 mmoles / 0.20 M = 497 mL (b) (2 marks) How much NaOH is required to completely convert the initial acetic acid solution to acetate? Moles acetic acid present = 0.250 L 0.40 M = 0.10 moles Volume NaOH required = moles acetic acid / conc. NaOH V (NaOH) = 0.10 moles / 0.20 M = 0.5 L...
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This note was uploaded on 03/25/2009 for the course CHEM 102 taught by Professor Drk during the Spring '09 term at University of Alberta.

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