- #1
Gaunt101
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Homework Statement
Determine the pH when 0.39 mol HOCl is added to sufficient water to make 1.000 L of solution. (Ka (HOCl) = 3.700 x 10-8)
Homework Equations
Ka = Kw/Kb
pH = -log(pKa) + 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.