Question

Cunoscand constanta de aciditate a acidului acetic Ka=2x10^-5mol/L, atunci pentru ionul acetat Kb este:
A. 4x10^-11mol/L
B. 15x10^-5mol/L
C. 75x10^-8mol/L
D. 5x10^-10mol/L
E. 25x10^-14mol/L

As dori si rezolvarea, nu doar raspunsul. Dau coroana celui ce o rezolva!!

Answer 1

CH3COOH + H2O <==> CH3COO- + H3O+

Ka = [CH3COO-]x[H3O+] / [CH3COOH]

=>  [CH3COOH]xKa = [CH3COO-]x[H3O+]

=>  [CH3COOH] = [CH3COO-]x[H3O+]/Ka

CH3COO- + H2O <==> CH3COOH + HO-

Kb = [CH3COOH]x[HO-] / [CH3COO-]

=>  [CH3COO-]xKb = [CH3COOH]x[HO-]

=> [CH3COOH] = [CH3COO-]xKb/[HO-]

=>  [CH3COO-]x[H3O+]/Ka =  [CH3COO-]xKb/[HO-]

=> [H3O+]/Ka = Kb/[HO-]

=> KaKb = [H3O+]x{HO-] = 10la-14

=> Kb = 10la-14/Ka

= 10la-14/2x10la-5 = 5x10la-10 mol/L