Asked by Rahul
what is the OH- concentration of 0.08 M solution of CH3COONa
Answers
Answered by
DrBob222
Call CH3COONa = NaAc and HAc is acetic acid; i.e., CHCOOH. It's the Ac^- that is hydrolyzed.
.......Ac^- + HOH ==> HAc + OH^-
I....0.08..............0.....0
C......-x..............x.....x
E....0.08-x............x.....x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.08-x). Solve for x = OH^-
.......Ac^- + HOH ==> HAc + OH^-
I....0.08..............0.....0
C......-x..............x.....x
E....0.08-x............x.....x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.08-x). Solve for x = OH^-
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