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Calculate the concentration of H3O+ in 0.120M hippuric acid given the pKa is 3.62

Nov 22nd, 2014

Thank you for the opportunity to help you with your question!

First we need to write the chemical reaction that occurs when the acid dissociates. HC9H8NO3 ---> C9H8NO3- H We can write an equilibrium expression for this reaction, products/reactants. K = [C9H8NO3-] [H ] / [HC9H8NO3] Now, we need to think about what is happening during dissociation. We start out with a certain concentration of the acid. thus, when it dissociates it loses concentration. lets say it loses x amount of concentration. so if the initial concentration of the acid is 0.120M, then at equilibrium it will be 0.120 - x. Now, when it dissociates, the products gain concentration.


Since  pKa = -log pKa

Ka = 24 * 10 -5

HC9H8NO3 = H3O C9H8NO3 0.12-x x x Ka = x 2 /(0.12-x) x 2 24 *10 -5 x - 4.56* 10 -5 =0

After solving the equation x = 6.634*10 -3 M

Conc. of H30 =6.634*10 -3 M

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 13th, 2015

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