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I am having trouble with this one problem. It is a couple of steps but I am very confused.
The problem states that there is 25 mL of a propionic acid where the concentration is unknown and it was titrated with .104M KOH. I got the Ka=1.34*10^-5 for the propionic acid. It also tells you that the equivalence point was reached at 35 mL.
The things I am confused about finding are the pH of the original KOH solution and of the propionic acid. Also, how would I be able to find the number of moles of KOH and propionic acid before the equivalence point is reached and after it is reached?
Thanks for your help.
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The problem is answered with a very detailed step-by-step formulaic explanation with notes throughout for optimal understanding.
This is a classic example of a reaction of a strong base with a weak acid.
First, write out the reaction:
KOH + propionic acid --------> potassium propionate (the salt) + H2O
How many moles of KOH were required to reach the equivalence point?
(0.104 mol/L)(0.035L) = 0.00364 mol KOH
At equivalence, there must be stoichiometrically equivalent amounts of acid and base that have reacted. Therefore, the number of moles of propionic acid present must be 0.00364 mol.
We can find out the pH of the starting propionic acid solution rather simply at this point. How do we do this? We do it by first determining the concentration of the starting solution.
0.00364 mol / 0.025 L = 0.15 M
How do we determine the pH of a solution of a weak acid?
Let's write out the generic ionization equilibrium for a weak acid in water, where HA is propionic acid and A- is propionate, the conjugate base.
HA + H2O <----> H3O+ ...
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