Asked by Diana
A .35M solution of a newly synthesized acid, HX, has [H+] of 4.1 x 10^-2. What is the value of Ka for this acid?
Answers
Answered by
DrBob222
...............HX ==> H^+ + X^-
initial.....(0.35M)..(0)....(0)
change........ -x.....x......x
final.......(0.35-x)..x......x
Ka = (H^+)(X^-)/(HX)
The problem tells you that x (H^+) = 4.1E-2. Substitute into the Ka expression above and solve for Ka.
initial.....(0.35M)..(0)....(0)
change........ -x.....x......x
final.......(0.35-x)..x......x
Ka = (H^+)(X^-)/(HX)
The problem tells you that x (H^+) = 4.1E-2. Substitute into the Ka expression above and solve for Ka.
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