Moderators: Chem_Mod, Chem_Admin

Patrick Wilson 2B
Posts: 22
Joined: Wed Sep 21, 2016 3:00 pm


Postby Patrick Wilson 2B » Sun Nov 27, 2016 2:31 pm

In class Dr. Lavelle talks about how much CH3COO (-) dissociates as a product when CH3COOH reacts with water. How do we calculate this dissociation value??

Posts: 22
Joined: Fri Jul 22, 2016 3:00 am

Re: Dissociation

Postby Hung_Sabrina_3N » Mon Nov 28, 2016 11:26 am

CH3COOH + H2O -> CH3COO- + H3O+
Is the equation for this dissociation. The Ka value will either be given or you have to find it.

You can find the amount of the product that is dissociated using the Ka value that would be given in this case it is 1.8× 10–5 M. You will use [CH3COO-][H3O+]/[CH3COOH]=Ka and use the initial molarity of CH3COOH to calculate for [CH3COO-][H3O+].

If you need to find Ka the concentrations of the reaction at equilibrium should be given then you just use [CH3COO-][H3O+]/[CH3COOH] to find Ka.

Return to “Conjugate Acids & Bases”

Who is online

Users browsing this forum: No registered users and 1 guest