A solution of household bleach contains 5.25% sodium hypochlorite, NaOCl, by mass. Assuming that the density of bleach is the same as water, calculate the volume of household bleach that should be diluted with water to make 500.0 mL of a pH = 10.10 solution.

5 Answers

5.25% NaOCl by mass means 5.25 g NaOCl/100 g soln and if the density is 1.00 g/mL (as is water) then molarity NaOCl solution is (5.25/74.44) mols NaOCl in 0.1L; therefore, M = 0.075/0.1L = estimated 0.71M. For pH = 10.10 is pOH 14-10.10 = 3.90 and since pOH = -log(OH^-) then (OH^-) = about 1.3E-4 but you need to do that more accurately.

OCl^- hydrolyzes in H2O as

……..OCl^- + HOH ==> HOCl + OH^-

I……..x…………..0……0

C…-1.3E-4………..1.3E-4..1.3E-4

E….x-1.34E-4……..1.3E-4..1.3E-4

OCl^- hydrolyzes in H2O as

……..OCl^- + HOH ==> HOCl + OH^-

I……..x…………..0……0

C…-1.3E-4………..1.3E-4..1.3E-4

E….x-1.34E-4……..1.3E-4..1.3E-4

Kb for OCl = (Kw/Ka for HOCl) = (1.3E-4)^2/(x-1.3E-4). You know Kw and Ka, solve for x = (OCl^-) needed. That comes out to approximately 0.05 but you should confirm that. I just did a quickie on the calculator.

Then c1v1 = c2v2

c = concn

v = volume

0.71*v1 = 0.05*500 mL

Solve for v1.

Please login or Register to submit your answer