Question

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

Answer 1

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 = ?