# Concentration of hypochlorous acid and hypochlorite

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.

• Note that chlorine is Cl, not "CL." I'm unclear on what you mean by "total pH of Cl." – Dissenter Aug 13 '14 at 16:15
• @Dissenter Yes oops I meant Cl, and total concentration of Cl. I will change it – user3325170 Aug 13 '14 at 16:23
• edit your question using latex – RE60K Aug 13 '14 at 17:52

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}