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Title: The acid-dissociation constant at 25.0 °C for hypochlorous acid (HClO) is 3.0 × 10-8? Post by: liz.goetzke on Sep 15, 2012 I would appreciate your unique working style to resolve this problem.
------------------------------------------------------ The acid-dissociation constant at 25.0 °C for hypochlorous acid (HClO) is 3.0 × 10-8. At equilibrium, the molarity of H3O+ in a 0.010 M solution of HClO is __________. a. 1.7 × 10-5 b. 0.010 c. 5.8 × 10-10 d. 4.76 e. 2.00 Title: The acid-dissociation constant at 25.0 °C for hypochlorous acid (HClO) is 3.0 × 10-8? Post by: linda71730 on Sep 15, 2012 HClO <>H+ + ClO-
initial concentration 0.010.......0.......0 at equilibrium 0.010-x.....x.......x K = 3.0 10^-8 = (x)(x)/0.010-x x=0.0000173 M pH = -log 0.0000173= 4.76 Title: The acid-dissociation constant at 25.0 °C for hypochlorous acid (HClO) is 3.0 × 10-8? Post by: fishnchips on Sep 16, 2012 1) The reaction of dissociation for hypochlorous acid is:
HClO + H2O <--------> H3O+ + ClO- 2) The acid-dissociation constant is then defined as follows: Ka = [H3O+][ClO-] / [HClO] we know that: Ka = 3.0 x 10-8 [H3O+] = x [HClO] = 0.010 M [ClO-] = x (according to the reaction above acid disociates equimolarly in both species: the hydronium and the hypochlorite ion) 3) We can substitute: 3.0 x 10-8 = x² / 0.010M we solve for x: x = 1.73 x 10-5 M That is the H3O+ concentration at equilibrium (option a) !!! Good luck! |