##### J. W. Hill, R. H. Petrucci, T. W. McCreary, & S. S. Perry General Chemistry 4th edition

Problem 15.67

For a solution that is 0.080 M NaOCl, (a) write an equation for the hydrolysis that occurs, and determine (b) the equilibrium constant for this hydrolysis and (c) the pH.

(a) NaOCl is a soluble salt, so NaOCl(aq) Na+(aq) + OCl(aq)

Na+ is the conjugate acid of a strong base, NaOH, so does not hydrolyze, i.e.

Na+(aq) + H2O(l) No Reaction

OCl is the conjugate base of a weak acid, HOCl, so does hydrolyze, i.e.

OCl(aq) + H2O(l)HOCl(aq) + OH(aq)

The solution will be basic.

(b) The equilibrium constant is found from the hydrolysis reaction

OCl(aq) + H2O(l)HOCl(aq) + OH(aq)

and has the form of a Kb, which must be found from the Ka of the conjugate acid HOCl:

assuming 25 oC.

(Remember that this relationship is only true for conjugate acid–base pairs.)

(c) Use the hydrolysis reaction that generates hydroxide, write mass action expression for the equilibrium constant, set up a table of concentrations, use this to find the pOH, and then find the pH.
 OCl–(aq) + H2O(l) HOCl(aq) + OH–(aq) Initial 0.080 0 0 Change – x + x + x Equilibrium 0.080 – x x x

Approximate? 0.080/3.4×10–7 = 240000 >> 100, so yes.

x = [OH] = 1.6×10–4 M

pOH = – log[OH] = – log[1.6×10–4] = 3.78

pH = 14.00 – pOH = 14.00 – 3.78 = 10.22