Oh but they do IF you handle Kh right. The hydrolysis constant = Kh = Kb = (Kw/Ka) or in the case of acids, Kh = Ka = (Kw/Kb). Said another way, the hydrolysis constant, Kh, for ClO^- is just Kb for ClO^-. What the author of the problem has done is s/he has already divided Kw/Ka and given you the answer of 2.86E-7 = Kh. Look in you text for Ka for HClO. My old book gives 3.5E-8 so
Kw/Ka = 1E-14/3.5E-8 = 2.857E-7 which rounds to 2.86E-7. Voila.
What is the molarity of OH- in a 7.83×10-3 M NaClO solution that hydrolyzes according to the equation.
ClO-(aq) + H2O(l) = OH-(aq) + HClO(aq)
The information they give is that constants Kh=2.86E-7
I know how to solve the problem when the constant Kw is involved (1E-14) i tried using those steps but they don't work with Kh..
1 answer
1 answer