Asked by devon
a solution of 2.5 M weak acid is .52% ionized. what is the Ka value of this acid?
ive been stuck on this one for a long time and cant figure out where to even start...if someone could walk me through this itd be great! thanks
ive been stuck on this one for a long time and cant figure out where to even start...if someone could walk me through this itd be great! thanks
Answers
Answered by
DrBob222
If the solution is 0.52M and it is 0.52% ionized, then the ions are 0.52% of 2.5M; therefore, (H^+) = (A^-) = 0.0052*2.5 = about 0.013M
......HA ==> H^+ + A^-
So plug thes numbers into the Ka expression and solve for Ka.
(H^+) = 0.013M
(A^-) = 0.013M
(HA) = 2.5-0.013 = ?
......HA ==> H^+ + A^-
So plug thes numbers into the Ka expression and solve for Ka.
(H^+) = 0.013M
(A^-) = 0.013M
(HA) = 2.5-0.013 = ?
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