February 19, 2017

Homework Help: chem

Posted by tomi on Thursday, April 16, 2009 at 2:09pm.

Calculate [H3O] in the following solutions.
5.15x10-2M HCl and 7.62x10-2M NaC2H3O2


HCl + NaC2H3O2. It appears there is an excess of H^+ which will react with NaC2H3O2 to form HC2H3O2. That gives you a buffer of C2H3O2 and HC2H3O2 and you use the H-H equation to solve for pH.
Post your work if you get stuck.

i used the H-H equation.
pKa=4.74, conjugate base= 7.62x10-2, Acid=5.15X10-2. i got a pH 0f 4.91 and H30 turned out to be 1.2X10-5.
Is this right?

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