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.

Question

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.

Answer 1

To calculate the Ka of the monoprotic acid, we need to use the information given, primarily the volume of NaOH solution added at the equivalence point and the pH of the solution after adding 10.0 ml of NaOH.

Let's break down the steps to determine the Ka:

Step 1: Calculate the concentration of OH- ions at the equivalence point.
Since NaOH is a strong base, it dissociates completely in water. Therefore, at the equivalence point, the moles of OH- ions added will be equal to the moles of acid initially present.

Given that 10.0 ml of NaOH was added, we can convert it to moles using the molarity of NaOH (which is unknown) and the volume.

Step 2: Calculate the concentration of H+ ions at the equivalence point.
The reaction at the equivalence point is as follows:
Acid + NaOH -> Salt + H2O

Since we added an equivalent amount of NaOH, the concentration of H+ ions is equivalent to the concentration of OH- ions.

Step 3: Calculate the concentration of H+ ions at the start of the titration.
Since the pH after adding 10.0 ml of NaOH was measured to be 5.0, we can use the pH to calculate the H+ ion concentration. pH is defined as the negative logarithm of the H+ ion concentration.

So, [H+] = 10^(-pH) = 10^(-5.0)

Step 4: Calculate the initial concentration of the acid (HA).
At the start of the titration, the concentration of the acid (HA) is equal to the concentration of H+ ions, which we calculated in Step 3.

Step 5: Calculate the moles of acid (HA) at the start.
The moles of acid (HA) can be calculated by multiplying the initial concentration of the acid (HA) by the volume of water in which it was dissolved. Since the volume of water is unknown, we'll leave it as a variable.

Step 6: Calculate the moles of acid (HA) at the equivalence point.
The moles of acid (HA) at the equivalence point are equal to the moles of OH- ions added at the equivalence point, which we calculated in Step 1.

Step 7: Calculate the initial concentration of the acid (HA) using the moles of acid (HA) at the equivalence point obtained in Step 6 and the volume of water used.

Step 8: Calculate the Ka using the concentrations of the acid (HA) at the start and at the equivalence point.
The Ka can be calculated by dividing the concentration of the salt by the concentration of the acid (HA) at the start of the titration.

Since we have calculated the necessary values, we can now substitute them into the equation and calculate the Ka.

Please note that we don't have all the necessary information to perform these calculations, as we don't know the molarity of NaOH or the volume of water used to dissolve the acid.