Acid/Base Problem

[Radagast]

A chemist dissolved 25.0g of CH3COOH in enough water to make 1.0L of solution.

What is the [H+]?

What is the [CH3COOH]?

 

Sounds easy, but it's giving me problems (mainly because I haven't worked with organic compound questions all that much). Any hints from you guys as to what to do?

 

So far I have converted the 25g to moles, and when I attempted putting my answer in the Kw = [H+][OH-] equation I got the wrong answer...

[jdurg]

CH3COOH is acetic acid. What you need to do is convert the mass into moles to get molarity, and use the Ka equation for acetic acid. (Ka = ([H+][CH3COO-])/[CH3COOH]). With that equation you then figure out the concentrations using the Ka value and the I.C.E. table. (Concentrations at Initial, change in Conc., and Concentrations at Equllibrium). You shouldn't have too much trouble with this one.

[P_Rog]

CH3COOH is a weak acid i believe, so it only partially dissociates. so turn 25 grams into moles, then divide by liters (1) to give you molarity, which is concentration. so...

 

CH3OOH --> CH300- + H+

(Molarity)..........+X......+X

M-X...................X........X

 

X^2/(M-X) = Ka which is the equilibrium constant for this reaction, which should be given or you'll have to look up.

 

X = [H+] and Molarity = [CH3OOH]

 

pretty sure that's how it's done

[Radagast]

k, got it now. Thanks guys