# chemistry

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What is the acid dissociation constant for a 0.010 M solution of the unkown acid HX that has a pH of 2.70 at equilibrium?

• chemistry -

pH = 2.70 = -log(H^+)
(H^+) = 0.002M

...........HX ==> H^+ + X^-
I.......0.010.....0......0
C..........-y.....y......y
E........0.010-y...y.....y

Ka = (H^+)(X^-)/(HX)
y = 0.002 from the pH calculation above. Substitute and solve for Ka.

• chemistry -

Thanks for the help, but I keep getting 4*10^-4 and not 5*10^-4 which is what I'm told the correct answer is. :S

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