Concentration of hypochlorous acid and hypochlorite
You really overcomplicate things (as I did initially, because I thought you were talking about the dissociation of $\ce{Cl2}$ gas in water and the subsequent reactions).

Here are the basic pieces of information needed:


Reaction equation
Equilibrium equation
Mass balance equation


Reaction Equation

The reaction is a simple acid-base equilibrium reaction:
$$ \ce{HOCl &lt;=&gt;[K_\text{a}] H+ + OCl-} $$

Equilibrium Equation

The equilibrium equation follows easily from the reaction equation:
$$ K_\text{a} = \frac{[\ce{H+}][\ce{OCl-}]}{[\ce{HOCl}]} $$

Mass Balance Equation

The total amount of chlorine is calculated as follows:
$$ [\ce{Cl}]_\text{tot} = [\ce{HOCl}] + [\ce{OCl-}]$$

From the reaction equation also follows
$$ [\ce{OCl-}] = [\ce{H+}] $$

The Solution

Since the concentration of protons is known, we can easily calculate the concentration of the hypochlorite anion:
$$ [\ce{OCl-}] = [\ce{H+}] = 10^{-\text{pH}} = 10^{-8} $$

Since we know the total concentration, simply reforming the mass balance gives us the concentration (strictly speaking, the activity) of the hypochlorous acid:
$$ [\ce{HOCl}] = [\ce{Cl}]_\text{tot} - [\ce{OCl-}] = 0.1 - 10^{-8} = 0.09999999 \approx 0.1 $$

Slap on some units and we're good to go!

$$ \begin{align}
c(\ce{HOCl}) &amp;\approx 0.1~\mathrm{mol\, L^{-1}} \\
c(\ce{OCl-}) &amp;= 10^{-8} ~\mathrm{mol\, L^{-1}}
\end{align}$$