Posted by Summer on Sunday, February 24, 2013 at 9:19pm.
A very old and tired , grey haired AP Chem instructor wanted to determine the Ka of an unlabelled monoprotic acid in his stockroom. He dissolved an unknown amount of acid in an unknown amount of water and proceeded to titrate the sample with a solution of NaOH of unknown molarity. After adding 10.0 ml of NaOH, he uttered that famous first order expletive “Oh, Michigan State”. He stopped and measured the pH of the solution at that point and found pH= 5.0. He continued to add NaOH until he realized he didn’t add phenolphthalein to the solution. He added 3 drops and the solution remained colorless. He continued the titration and found the equivalence point to be 32.22 ml of the NaOH solution. Can our intrepid hero calculate the Ka? Hint: he can. So calculate the Ka. Show all of your work.
Chemistry - Summer, Sunday, February 24, 2013 at 10:16pm
At the first point where he stopped and measured the pH, the H+ ion concentration is 1*10^-5 and the OH- ion concentration is 1*10^-9. This is as far as I've gotten, I can't figure out what to do with this information.
Chemistry - DrBob222, Monday, February 25, 2013 at 12:35am
I figure at 10 mL, where the pH = 5, at that point some of the "base"(the anion) has been formed so you should be able to use the Henderson-Hasselbalch equation and calculate pKa.
pH = pKa + log(base)/(acid)
5.0 = pKa + log (base)/(acid)
At 10 mL it seems to me that the base formed is 10/32.22 How much of the acid is left at this point? It has neutralized 10 mL so it has another 22.22 mL to go so that fraction is 22.22/32.22. Substitute into the HH equation and solve for pKa.
Check my thinking.
Chemistry - Summer, Monday, February 25, 2013 at 6:24pm
I plugged those numbers into the HH equation and came up with 2.22*10^-5 for the Ka.
Does that sound right?
Chemistry - Summer, Monday, February 25, 2013 at 9:39pm
It's actually 4.5*10^-6.
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