Finding Ka of Monoprotic Acid using information from titration

[Orims]

1. This problem has been slowly eating me for the past 2 hours. I've done everything I can but I can't seem to answer it. The pH at the equivalence point in the titration of 100 mL of a 0.1M monoprotic acid solution with a 0.1M strong base solution is 8.12 at 25 degrees C. What is the Ka of the acid?



I know the Ka of the acid can be found using pH= pKa + log ([A-]/[HA]) but only if it is before the equivalence point.



3. I found out that the equivalence point was when 100 mL of strong base have been added (same molarity as weak acid, so equivalence point = double the volume of initial acid solution). This would make the half equivalence point 150 mL.

 

[Borek]

What if I will reword the question for you: what is Ka of a weak acid, if 0.05M solution of its salt has pH 8.12?

 

[Orims]

Thank you! I finally got it!

I used the fact that [OH-] ~= sqroot(Kb X [A-]), so I got [OH] from pOH and used the fact that since HA and NaOH were in equal volumes and concentrations, the [A-] would be half the concentration of HA. I got Ka from Kb (Ka= Kw/Kb)

In the end I got a Ka of about 1.45x10 to the -5 and my book says the Ka would be 1.5x10 to the -5 (they round up).

Again, thank you!

 

[Borek]

[OH-] ~= sqroot(Kb X [A-])

You should check if conditions needed to use this equation are meet. This is only approximation and it is not always correct.

Calculation of pH of a weak acid/base.

 

[DeeNos]

I understand the frustration of not being able to solve a problem after spending a significant amount of time on it. However, I would like to assure you that there is a solution to this problem.

Firstly, it is important to understand the concept of titration and how it can be used to determine the Ka of a monoprotic acid. In a titration, a strong base is gradually added to a solution of a weak acid until the equivalence point is reached. At the equivalence point, the number of moles of acid is equal to the number of moles of base added. This means that the concentration of the acid and its conjugate base are equal at the equivalence point.

In your case, the pH at the equivalence point is given as 8.12. This means that at the equivalence point, the concentration of the acid and its conjugate base are both 10^-8.12 M. Using the Henderson-Hasselbalch equation, we can calculate the pKa of the acid as 8.12.

pKa = pH + log([A-]/[HA])
8.12 = 8.12 + log([10^-8.12]/[10^-8.12])
8.12 = 8.12 + log(1)
8.12 = 8.12

Now, we can use the pKa value to calculate the Ka of the acid using the equation Ka = 10^-pKa.
Ka = 10^-8.12 = 6.59 x 10^-9

In conclusion, the Ka of the monoprotic acid is 6.59 x 10^-9. It is important to note that this value may vary slightly due to experimental errors and other factors. It is always a good practice to repeat the experiment and take multiple measurements to ensure accuracy.

I hope this explanation helps you understand the concept of titration and how it can be used to determine the Ka of a monoprotic acid. Keep up the good work and don't let this problem discourage you. As a scientist, it is important to persevere and keep trying until you find the solution.