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 determine the Ka of the unlabelled monoprotic acid, we'll need to use the given information about the titration process and pH measurements. Let's go step by step.

Step 1: Calculate the initial concentration of the unlabelled acid.
The instructor dissolved an unknown amount of acid in an unknown amount of water, so we don't have direct information about the concentration. However, we can assume that the solution was initially dilute, so we will start with a concentration of 0. Remember that the volume of the solution does not affect the acid concentration in this case since we're dealing with a monoprotic acid.

Step 2: Calculate the concentration of OH- ions at the equivalence point.
The equivalence point is reached when all the acid has reacted with the NaOH solution. Since the reaction is 1:1 (monoprotic), the moles of OH- ions added equals the moles of acid initially present. Using the volume of NaOH solution added at the equivalence point (32.22 ml, which is equivalent to 0.03222 L), we can calculate the moles of OH- ions added.

Step 3: Calculate the concentration of H+ (or H3O+) ions at the equivalence point.
Since the reaction is neutralization, at the equivalence point, the moles of OH- ions added will be equal to the moles of H+ (or H3O+) ions initially present. Therefore, using the volume and concentration of the unlabelled acid, we can calculate the concentration of H+ ions at the equivalence point.

Step 4: Calculate the pKa.
The pKa is the negative logarithm of the Ka, so we'll need to find the Ka to get the pKa. To do this, we'll use the pH measurement at the midpoint of the titration (when 10.0 ml of NaOH was added).

At the midpoint, the volume of NaOH added is half the volume at the equivalence point. So, the volume at the midpoint is 0.5 * 32.22 ml = 16.11 ml (equivalent to 0.01611 L).

Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])

At the midpoint, [A-] and [HA] are equal since half the acid has been titrated. Assuming the initial concentration of the acid is C (mol/L), at the midpoint it would be C/2 (mol/L).

Given that pH = 5.0, we can rearrange the equation:
5.0 = pKa + log(0.5C/C)

Simplifying:
5.0 = pKa + log(0.5)

Step 5: Calculate the Ka.
Using the equation from step 4, we can solve for pKa:
pKa = 5.0 - log(0.5)

Finally, to get the Ka, we take the antilog (inverse log) of the pKa:
Ka = 10^(-pKa)

By substituting the calculated pKa into this equation, we can find the value of Ka for the unlabelled monoprotic acid.

Note: Some assumptions have been made in this calculation, such as the dilute initial concentration of the acid and the absence of other acid-base reactions in the solution.