- #1

Gaunt101

- 8

- 0

## 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.