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.