Calculate the pH of a 0.490 M solution of NaOCl?

Question

HINT: find Kb from Ka of the conjugate acid

It would be greatly appreciated if you could show how you derived the answer.
Thanks!

Answer 1

Na+ is the salt of a strong base (NaOH) so its neutral. OCl- is the salt of a weak acid (HOCl) and that will react with water to make a basic solution as follows:
OCl- + H2O --> HOCl + OH-
Ka = 3.5E-8
Kb = E-14/3.8E-8 = 2.86E-7
Initial concentrations:
[OCl-] = 0.490 M
[HOCl] = 0 M
[OH-] = 0 M
Equilibrium concentrations:
[OCl-] = 0.490 - x
[HOCl] = x
[OH-] = x

Kb = [HOCl][OH-]/[OCl-]
2.86E-7 = [(x)(x)]/(0.490 - x)
since Kb is really small, assume that OCl- doesn't dissociate and the final concentration to be 0.490 M
2.86E-7 = (x^2)/0.49
x = 3.74E-4
[OH-] = x = 3.74E-4
pOH = -log(3.74E-4) = 3.43
pH = 14 - 3.43 = 10.57
the pH indicates that the solution is basic which verifies the prediction made in the reaction above

Answer 2

1) First of all, we need to know the Ka of the conjugate acid (in this case hypochlorous acid):

NaClO + H2O <----> HClO + Na(+) + OH-

pKa (HClO) = 7.497 (reported in tables)
as we know that:

pKa + pKb = pKw = 14

then:

pKb = 14 - 7.497 = 6.503
Kb = 10^-(6.503) = 3.14 x 10^-7

2) We know that, Kb = [HClO-][OH-] / [NaClO]
that means that in equilibrium: Kb = x² / (0.490 - x)
3.14 x 10^-7 = x² / (0.490 - x)

That is a quadratic equation we can solve for x:
x = 3.92 x 10^-4 (we only take the positive root)
Remember that x is the base concentracion (OH- concentration)

3) Finally we can compute the pH as:

pOH = -log (3.92 x 10^-4) = 3.406
pH = 14 - 3.406 = 10.59
That's it!

Hope it helps!