# What is K_a for HOCl, given that initially [HOCl]=0.08*mol*dm^3, and pH=2.85?

Apr 23, 2017

K_a=([H_3O^+][""^(-)OCl])/([HOCl])=2.54xx10^-5.........

#### Explanation:

The $\text{acid dissociation constant,}$ ${K}_{a}$, measures the particular acid dissociation............

HOCl(aq) + H_2O rightleftharpoonsH_3O^(+) + ""^(-)OCl

Where K_a=([H_3O^+][""^(-)OCl])/([HOCl])

We are given that the initial concentration was $0.08 \cdot m o l \cdot {\mathrm{dm}}^{-} 3$, and that $p H = 2.85$. But we know that $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$.

And thus $\left[{H}_{3} {O}^{+}\right] = {10}^{- 2.85} \cdot m o l \cdot {L}^{-} 1$.

But, by stoichiometry [""^(-)OCl]=10^(-2.85)*mol*L^-1.

And, thus......................

$\left[H O C l\right] = \left(0.08 - {10}^{- 2.85}\right) \cdot m o l \cdot {L}^{-} 1 = \ldots \ldots \ldots \ldots \ldots$

$= 0.0786 \cdot m o l \cdot {L}^{-} 1$.

And thus we can calculate ${K}_{a}$ sans approximation, i.e.

${K}_{a} = \frac{{\left({10}^{- 2.85}\right)}^{2}}{0.0786} = 2.54 \times {10}^{-} 5$.