HA ----> A^- + H^+
Ka=[A^-][H^+]/[HA]
....HA........A^-....H^+
I ..0.22 ......0......0
C..-x..........x.......x
E.0.22-x....x.......x
Ka=[x][x]/[0.22-x]
Let x=0.22*(0.209)
Solve for x:
A weak acid (HA) is found to ionize 20.9%
when mixed at a concentration of 0.22 mol/L.
What is the value of Ka for this acid?
5 answers
I believe the problem asks for Ka.
it is wrong
Typed something out of habit.
plug in the value for x and solve Ka:
I apologize about that.
plug in the value for x and solve Ka:
I apologize about that.
I meant to say solve for Ka and not x.