Calculate the pH of a 0.460 M solution of NaOCl?

Question

HINT: find Kb from Ka of the conjugate acid

Answer 1

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!

Answer 2

So, you know the (or can find in your text) that Ka times Kb = Kw = 1 X 10^-14.

So, you can find the Ka for HOCl, and use that to calculate Kb.

Now, the base OCl- will react with water like this:

OCl- + H2O --> HOCl + OH-

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

The concentrations of HOCl and OH- will be equal, so let them both equal x. The [OCl-] will properly equal 0.460-x, but you might be able to ignore that x, and let [OCL-]=0.460.

Plug all those into the equation for Kb and solve for x which is equal to the [OH-].

Are you OK getting from OH- concentration to hydronium ion concentration and to pH?

Answer 3

In water, NaOCl will dissociate according to the following:

NaOCl ----> Na+ + ClO-

Since water is made of HOH, meaning H+ and OH- in equal concentraion, the pH of water is 7

but, the H+ will react with ClO- to form this weak acid HClO

since it is weak acid, the H+ will bond with ClO- and not dissociate, thus it will decrease the concentration of H+, but at the same time incerasing the concentration of OH-

now because there is more OH-, the solution is going to be basic. to find thepH, you first need to find the Ka value of HClO

Ka=2.9 x 10 - 8

HClO <----> H+ + ClO-
Initial: .460M <-> 0 + 0
Change: -x <-> +x + +x
______________________________
Equilibrium: .460-x <-> x + x

now:
Ka=concentration of products/reactants
2.9e-8=x^2/(.460-x)
solve for x
x=6.14e-9
now that is the concentration of H+
to find the pH
-log([H+])=pH
-log(6.14*10^-9)=8.21

Answer 4

Hint : What is Ka of conjugate acid (HOCl) ??????????