I need help with a chem lab. I need to find the ka of 10 ml ~.1 acetic acid diluted with 40 ml H2O, and titrated with .075M NaOH. I know the equivalence point is supposed to be about 13.33 ml (M1V1=M2V2) so my half-way point should be about 6.67 ml. Here's my problem: my readings for ml go by .5, so I don't have a 6.67ml reading, only a 6.5 and 7.0 ml. Same with the equivalence, I've got 13.0 and 13.5, but not 13.33. Any help will be much appreciated.
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Extrapolate between the two values that you do have.
For example, say you have the following:
At 6.50 mL, pH = 4.62
At 7.00 mL, pH = 4.85
(6.67 mL - 6.50 mL) / (7.00 mL - 6.50 mL) = 0.17 / 0.50 = 0.34 . . .6.67 mL is 34%of the way between 6.50 mL and 7.00 mL. Now find the pH that is 34% of the way between 4.62 and 4.85.
4.85 - 4.62 = 0.23; 0.34 x 0.23 = 0.08; pH at 6.67 mL = pH at 6.50 mL + 0.08 = 4.62 + 0.08 = 4.70
pKa = 4.70 and Ka = 10^-pKa = 10^-4.70 = 2.0 x 10^-5
For example, say you have the following:
At 6.50 mL, pH = 4.62
At 7.00 mL, pH = 4.85
(6.67 mL - 6.50 mL) / (7.00 mL - 6.50 mL) = 0.17 / 0.50 = 0.34 . . .6.67 mL is 34%of the way between 6.50 mL and 7.00 mL. Now find the pH that is 34% of the way between 4.62 and 4.85.
4.85 - 4.62 = 0.23; 0.34 x 0.23 = 0.08; pH at 6.67 mL = pH at 6.50 mL + 0.08 = 4.62 + 0.08 = 4.70
pKa = 4.70 and Ka = 10^-pKa = 10^-4.70 = 2.0 x 10^-5