Ka problem?

[H+] is 2.51E-3 in a 0.35 M solution of acetic acid (CH3COOH). What is the Ka?

please show your steps, or explain to me how you did it

thanks!!

Answer:
K is the ratio of products to reactants. The hydrolysis reaction of acetic acid is
CH3COOH + H2O -> CH3COO- + H3O+

Water doesn't count in your K concentrations since its concentration doesn't change.

So, Ka = [CH3COO-][H3O+]/[CH3COOH]

Since you create one H3O+ for every CH3COO- you create by deprotonating the acid, those concentrations are equal. The acids dissociates so little that you needn't worry about subtracting any value from the acetic acid concentration you started with.

So the first answer is good: (2.51E-3)^2 divided by .35 is your Ka. Just thought you needed a little more explanation so you can get the next one on your own.

Sqaure the H+ ion concentration, and divide it by the concentration of the acid.

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