solve Ka for HX => H^+ + X^- given 0.1M LiX at pH=8.9

Answer 1

LiX(aq) ==> Li^+(aq) + X^-(aq)

X^- is a base and hydrolyzes in water to
X^- + HOH ==> HX + OH^-
Make an ICE chart and substitute into the following:
Kb = (Kw/Ka) = (HX)(OH^-)/(X^-)
You know pH, convert to pOH, then to OH^-. HX = OH^-. You know Kw. The only unknown is Ka. Solve for that. Post your work if you get stuck.

Answer 2

Is the X^- in the ICE going to be 0.1 from the initial o.1M LiX?

Answer 3

See above.

Answer 4

i cant understand please elaborate