A solution of household bleach contains 5.25% sodium hypochlorite, NaOCl, by mas

Question

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 = 10.26 solution.

my teacher did this question like this:

500 x .00525 = 2.625 g naocl

2.625/74.5 = .035 mol     .035/.5= .07

Ka hclo3= 3.8x10^-8    Kb=1x10^-14 / 3.8x10^-8 = 2.63x10^-7

14-10.26= 3.74          10^-3.74= 1.82x10^-4

2.63x10^-7 = (1.82x10^-4)^2 / x                  x= .126mol

.126/.5L = .252 mol

.00525/ 74.5 = 7.05x10^-5

7.05x10^-5 / .1L = 7.05x10^-4

.252/ 7.05x10^-4 = 357.44 ml

I understand how she did this but I dont get where she got the Ka hclo3= 3.8x10^-8 from

She assigned this same problem but the ph is 10.10 and Im just trying to find out where that Ka hclo3 came from so i can solve the rest of the problem

Explanation / Answer

you understand total solution that's good . you only problem with 3.8x10^-8 value.

that is not HClO3 . it is HClO

HClO weak acid . it s Ka value from standard table = 3.8 x 10^-8 .   you can see any standard table you will find.

why we are taking HOCl Ka value . because here NaOCl is given. it is the salt of strong base weak acid . so for the hydrolysis of this salt we need to take Ka

NaOCl ----------------------> Na+   + OCl-

here OCl- involve in hydrolysis process

OCl-   + H2O <-------------------------> HOCl   + OH-

so it is base hydrolysis . we know Ka of HOCl we can calculate Kb of OCl-

Kb = [HOCl] [OH-] / [OCl-]

Ka x Kb = Kw

Kb = Kw / Ka = 1.0 x 10^-14 / 3.8 x 10^-8

Kb = 2.63x10^-7

i think you got it now.

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