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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