Calculate the pH of a 0.500 M solution of NaOCl?

Question

HINT: find Kb from Ka of the conjugate acid

Answer 1

Ka HOCl is 3.5x10^-8
pKa HOCl = -log(3.5 x 10^-8)
=7.4559

pKb = 14 - 7.4559 = 6.54406
Kb HOCl = 10^-6.54406
Kb =2.85759 x 10^-7

NaOCl(aq) + H2O <---> NaOH(aq) + HOCl(aq)

Kb = [product] / [reactants]
Kb = [OH][HNaOCl] / [NaOCl]
Kb = x^2 / 0.500
2.85759 x 10^-7 = x^2 / 0.500
x^2 = 0.500(2.85759 x 10^-7)
x^2 = 1.43795 x 10^-7
x = 3.792 x 10^-4

-log(3.792 x 10^-4)=3.4211

which doesn't make any sense since it is suppose to make a weak base. I thin the Ka value is wrong. The rest is right I think.

Answer 2

when the molecule of a weak base NaOCl is dissolved in water, it dissociates to form: NaOCl =Na+ + OCl- . If this dissociation is complete then we have:
OCl- + H2O =HOCl + OH-
Therefore the equillibrium constant for this reaction, based on equillibrium concentrations is (Kc) is given as follows:

Kb=[OH-][HOCL]/[H2O][OCl-]

since the concentation of water[H2O] is constant, it is incorporated into the equillibrium constant to give (Kb)

Kb=[OH-][HOCL]/[OCl-]
(remember all these are equilllibrium constrations of each specie involved)
when we look on Kb table we are unlucky to get the value of (Kb), but don't worry, we can use the Ka, The conjugate acid dissociation constant, instead.

first, you have to know that : {Ka *Kb = 10 ^-14 }

The conjugate acid for NaOCl is what what we get when it dissociates( look at the equation above for its dissociation in water) and is {HOCl}

on the table for (Ka ) we get the value of 2.9*10^-8

and use the relationship: Ka*Kb=10^-14
therefore Kb=(10^-14)/(Ka)
=(10^-14)/(2.9*10^-8)
=3.45*10^-7
for each OCl- which melts we have one HOCl and one OH- formed. Therefore their concentrations are equal, and we have
[HOCl ]=[ OH- ] which can be written as [ OH- ] ^2 for our convenience.
therefore we now have:
Kb=[OH-]^2/[OCl-]
therefore Kb/ [OCl-]=[OH-]^2

and [OH-]=√Kb/ [OCl-]
as we have all values :
concentration of OCl-=0.5 ( given in question)

[OH-]=-√(3.45*10^-7)/(0.500)
=8.3*10^-4
but we know that Kw(ionic product of water) is as follows:

Kw=[OH-]*[H+]=1*10^-14 and is a constant.
by using it
we have [OH-] and we need [H+] Therefore

[H+]=(1*10^-14 )/[OH-]
=(1*10^-14 )/ (8.3*10^-4)
=1.2810^-11

Therefore the pH is now easy to find:

pH=-log[H+]
=-log(1.2810^-11)

=10.9

quite basic solution, as we expected. If we could have got a value less than 7 , something would have gone wrong, because it would indicate that NaOCl is acidic, which would be totally wrong.

This is as far as I can explain, and I hope you are helped.

Answer 3

The equation governing this is
OCl- = OH- + HOCl
We expect all the salt to dissolve.

One can show that:

[OH-][HOCl] / [OCl-] = kw/ka
If x amount of salt dissociates to OH- and free acid and x<< OCl-,
x^2/0.5 = kw/ka, which is too hard to solve.