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Calculating Ka Value of Unknown Acid?
Alright, so I need to calculate the Ka value of an unknown acid using the amount of NaOH (0.1002 M) titrated and the pH of the solution.
The equivalence point is at 23.45 mL NaOH, with a pH of 10.68
Amount of acid was 25 mL, unknown concentration.
Any help would be appreciated, and I will def pick a best answer. Thanks.
2 Answers
- ChemTeamLv 79 years agoFavorite Answer
At the equivalence point, this reaction is the one to focus on:
A- + H2O <==> HA + OH-
Since pH = 10.68, the pOH = 3.32
[OH-] = 10^-pOH = 10^-3.32 = 4.7863 x 10^-4 M (this is also the concentration of HA)
The concentration of A- is this:
(0.1002 mol/L) (0.02345 L) = 0.00234969 mol
0.00234969 mol / 0.025 L = 0.0939876 M
Kb = [(4.7863 x 10^-4) (4.7863 x 10^-4)] / 0.0939876
Kb = 2.4374 x 10^-6
KaKb = Kw
(Ka) (2.4374 x 10^-6) = 1.00 x 10^-14
Ka = 4.1027 x 10^-9 (round off as you see fit)
Source(s): ChemTeam - Anonymous5 years ago
Does an answer key say it is a)? you can find Ka by taking the inverse log of pH or 10^-2 which leads to answer c.... I do not believe the 1.0M should come into the equation.. but I could be wrong. I am using inverse log b/c pH= -log(Ka) So the M does come into play.... :) Good to know!
