Asked by Larry
solve Ka for HX => H^+ + X^- given 0.1M LiX at pH=8.9
Answers
Answered by
DrBob222
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.
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.
Answered by
Larry
Is the X^- in the ICE going to be 0.1 from the initial o.1M LiX?
Answered by
DrBob222
See above.
Answered by
christina
i cant understand please elaborate
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