Solving pH for HOCl in a 1 L Solution

[Gaunt101]

Homework Statement

Determine the pH when 0.39 mol HOCl is added to sufficient water to make 1.000 L of solution. (K_a (HOCl) = 3.700 x 10-8)

Homework Equations

K_a = K_w/K_b
pH = -log(pK_a) + log(acid/base)

The Attempt at a Solution

For this question first we need the concentration: c= n/v hence the concentration of HOCL is 0.39M.

At this point if you write the balanced equation I think it's HOCl + H2O <---> H3O + ClO
so there is a positive increase in H3O and ClO, hence:

Ka = [H3O]⁺[ClO]^- / [HOCL]
by letting H3O be x, then both the concentrations of H3O and ClO will be x^2 since they increase the same.

So I figure Ka = x^2 / [HOCL]

x = 1.170 e-4

so the pH = -log(x) + log(acid/base)
pH = 3.932 + log(acid/base)

I'm not sure what values I'd put in for acid / base, and I'm not sure if what I've done so far is correct.

Thank you very much for your time!

Gaunt.

 

[Borek]

Check this page:

calculation of pH of a weak acid solution

or this page:

http://en.wikipedia.org/wiki/ICE_table

This a different approaches, but they lead to the same answers.

 

[Gaunt101]

Hey Borek,

thanks for the sites, but I"m a little confused. Chembuddy says "the sum of concentrations of all acid forms present in the solution must be identical to the concentration of acid added."

Does that mean for the log(acid/base), the acid value would be the Ka and the base just the Kb?

 

[Borek]

No, it is just a mass balance. HClO dissociates, so solution contains both HClO and ClO^-. In your case mass balance for hypochloric acid means that

[HClO] + [ClO^-] = 0.39M

--

 

[Gaunt101]

[HClO] + [ClO-] = 0.39M

So if you were to do log(acid / base), it would simply be 0.39 since all the sums of the acid must be equal to the original concentration? hence the answer is just 0.39 + 3.932?

 

[Borek]

No, you don't know what are concentrations of HClO and its conjugate base ClO^-, so you can't calculate their ratio and log of their ratio. You have to calculate them first. Simplest approach is to calculate in reverse - that is, calculate pH of the solution, then use it to calculate ratio of concentrations.

log(acid/base) is a way of dealing with buffers, not with solutions of pure acid.