Chemistry

posted by .

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 -

    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 -

    0mL|......10mL|.................32.22 mL|
    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 -

    I plugged those numbers into the HH equation and came up with 2.22*10^-5 for the Ka.
    Does that sound right?

  • Chemistry -

    It's actually 4.5*10^-6.
    5=pKa+log(10/22.22)

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chemistry

    A 0.125-g sample of monoprotic acid of unknown molar mass is dissolved in water and titrated with 0.1003 M NaOH. The endpoint is reached after adding 20.77 mL of base. What is the molar mass of the unknown acid?
  2. AP Chemistry

    There is an unknown amount of unlabelled monoprotic acid in an unknown amount of water titrated with a sample with a solution of NaOH of unknown molarity. After adding 10.0 mL of NaOH, the pH=5.0. The equivalence point is 32.22 mL …
  3. Chemistry

    A 0.5224g sample of an unknown monoprotic acid was titrated with 0.0998M of NaOH. The equivalence point of the titration occurs at 23.82 mL. Determine the molar mass of the unknown acid.
  4. AP Chemistry

    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 …
  5. Chemistry

    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 …
  6. ap chemistry

    A 0.456 gram sample of an unknown mono- protic acid (let’s call it HZ) was dissolved in some water (you pick the amount). Then the acidic solution was titrated to the equivalence point with 32.5 mL of 0.174 M KOH. What is the molecular …
  7. Chemistry

    At 20 degree celsius, Ka for an unknown monoprotic weak acid is 0.00088. Calculate the Delta G (in kJ/mol) for the reaction of this unknown acid with water. HA(aq) + H2O(l) <--> H3O+(aq) + A-(aq)
  8. chemistry

    A 125.0 mg sample of an unknown, monoprotic acid was dissolved in 100.0 mL of distilled water and titrated with a 0.050 M solution of NaOH. The pH of the solution was monitored throughout the titration, and the following data were …
  9. chemistry

    a 3.54 grams solid sample of an unknown monoprotic acid was dissolved in distilled water to produce a 47.0 mL solution at 25 degrees. This solution was then titrated with 0.2 M NaOH. The equivalence point was reached when 35.72 mL …
  10. chemistry help

    a 3.54 grams solid sample of an unknown monoprotic acid was dissolved in distilled water to produce a 47.0 mL solution at 25 degrees. This solution was then titrated with 0.2 M NaOH. The equivalence point was reached when 35.72 mL …

More Similar Questions