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Group Number Sodium hypochlorite solution, sold us
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Answer #1

pH of m = 6.5% ww of NaClO

assume:

V = 1 Liter solution

so

mass of solution = D*V = 1*1 = 1000 g/mL

of which

6.5% is NaClO --> 65 g of NaOCl

MW of NaOCl = 74.44 g/mol

mol = mass/MW = 65 /74.44 = 0.873 mol of NaOCl

recall that:

NAOCl --> NA+ + OCl-

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

so

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

and

Kb = Kw/Ka

from tables

Ka = 3.5*10^-8

Kw = 10^-14 constant always

so

Kb = (10^-14)/(3.5*10^-8) = 2.8571*10^-7

assumptions:

[HOCl ] = x= [OH-]

[OCl-] = M- x = 0.873 /1 - x = 0.873- x

2.8571*10^-7 = x*x/(0.873- x )

x = 4.986*10^-4

[OH-] = x= 4.986*10^-4

pOH = -log(OH) = -log(4.986*10^-4) = 3.302

p´H = 14 -pOH = 14-3.302

pH = 10.698

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