Finding Ka of Monoprotic Acid using information from titration

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Discussion Overview

The discussion revolves around calculating the acid dissociation constant (Ka) of a monoprotic acid using data obtained from a titration with a strong base. The focus includes the pH at the equivalence point and the relationships between concentrations of species involved in the titration.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant describes their struggle to determine the Ka of a monoprotic acid given the pH at the equivalence point of a titration.
  • Another participant rephrases the problem, suggesting a focus on the pH of a salt solution to derive Ka.
  • A participant shares their successful calculation of Ka using the relationship between hydroxide ion concentration and the base dissociation constant (Kb), along with the concentrations of the acid and its conjugate base.
  • Another participant cautions that the approximation used to derive [OH-] may not always be valid and suggests checking the conditions for its application.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the approximation used in the calculations, indicating that there is no consensus on the conditions under which it applies.

Contextual Notes

The discussion includes assumptions about the concentrations and conditions under which the calculations are made, as well as the potential limitations of the approximation for [OH-].

Orims
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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.
 
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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?
 
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!
 
Orims said:
[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.
 

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