Asked by kayla
A 0.500M solution of a weak acid, HX, is only partially ionized. The [H+] was found to be 4.02 x 10^-3 M. Find the dissociation constant for this acid.
Answers
Answered by
DrBob222
HX ==> H^+ + X^-
Ka = (H^+)(X^-)/(HX)
Set up an ICE chart, substitute for each component in the Ka expression and solve for Ka. I get 3.26 x 10^-3. Don't forget to subtract 4.02 x10^-3 from 0.500 M to obtain unionized (HX); i.e., it isn't 0.500, it is 0.500 - 0.00402.
Ka = (H^+)(X^-)/(HX)
Set up an ICE chart, substitute for each component in the Ka expression and solve for Ka. I get 3.26 x 10^-3. Don't forget to subtract 4.02 x10^-3 from 0.500 M to obtain unionized (HX); i.e., it isn't 0.500, it is 0.500 - 0.00402.
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