# chemistry

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Calculate [acetate] of 0.057 mol/L solution of acetic acid of Ka = 1.76x10^-5

i know to find Ka it's:
Ka= products/reactants
so i tried doing an ICE table with the equation to be:
CH3CO2 + H2O <--> CH3COOH + H
am i on the right track? if not...help please!

• chemistry -

Yes, you just stopped too soon.
Let's call acetic acid HAc.
HAc --> H^+ + Ac^-

Then, as you have it, Ka = (H^+)(Ac^-)/(HAc) = 1.76 x 10^-5.

Now you do the ICE chart.
(H^+) is x
(Ac^-) is x
(HAc) is 0.057 -x

I will leave the equation for you to solve. Please note, however, that while the x in 0.057-x USUALLT can be neglected, I don't think it can in this problem and you will need to solve the quadratic equation (or do it by successive approximations). The only real difference between this problem and most of the others like this one is that this one asks for Ac^- (instead of asking for pH etc) and all you need to do is to realize that acetate and H^+ are the same.

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