Concentration of hypochlorous acid and hypochlorite

Question

Given a total chlorine concentration $c(\ce{Cl})_\text{tot}= \pu{0.1 M}$, with $\mathrm{pH}(\ce{Cl}) = 8$ and $\mathrm{p}K_\mathrm{a}(\ce{HOCl}) = 7.5$, how would you find the concentration of $\ce{OCl-}$ and $\ce{HOCl}$ disinfectants?

So far I think that I should start this way:

$K_\mathrm{a} (\ce{HOCl}) = 10^{-7.5}=3.16\cdot 10^{-8}$

I wrote up the reactions:

$$\ce{H2O + 2e- + OCl <=> Cl- + 2OH-}$$

Where $\ce{OCl-}$ is the conjugate acid and $\ce{OH-}$ the conjugate base.

$$\ce{HOCl<=> OCl- + H+}$$

The conjugated acid is $\ce{HOCl}$, the conjugated base $\ce{OCl-}$.

Now I use $$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log\left(\frac{[\text{conj base}]}{[\text{conj acid}]}\right)$$

and/or for weak acids $$K_\mathrm{a}= [\text{products}]/[\text{reactants}]$$

But I'm not sure if this is correct and what values to use.

Answer 1

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:

1. Reaction equation
2. Equilibrium equation
3. Mass balance equation

Reaction Equation

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

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!

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