# Question #99897

Jun 4, 2015

The concentration of the hydronium ions will be equal to $8.8 \cdot {10}^{- 5} \text{M}$.

Hypochlorous acid, $H C l O$, is a weak acid that dissociates in aqueous solution to form hydronium cations, ${H}_{3} {O}^{+}$, and hypochlorite anions, $C l {O}^{-}$.

You can determine the concentration of the hydronium ions by using an ICE table for the following equilibrium reaction

$\text{ } H C l {O}_{\left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{3} {O}_{\left(a q\right)}^{+} + C l {O}_{\left(a q\right)}^{-}$
I......0.265......................................0.....................0
C.......(-x).......................................(+x).................(+x)
E.....0.265-x...................................x.....................x

The acid dissociation constant, ${K}_{a}$, will be equal to

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \cdot \left[C l {O}^{-}\right]}{\left[H C l O\right]} = \frac{x \cdot x}{0.265 - x} = {x}^{2} / \left(0.265 - x\right)$

Because ${K}_{a}$ has such a small value, you can approximate (0.265 - x) with x to get

${K}_{a} = {x}^{2} / 0.265 = 2.9 \cdot {10}^{- 8}$

This means that $x$ will be equal to

$x = \sqrt{0.265 \cdot 2.8 \cdot {10}^{- 8}} = 8.8 \cdot {10}^{- 5}$

Since $x$ is equal to the molarity of the hydronium ions, the answer will be

$\left[{H}_{3} {O}^{+}\right] = \textcolor{g r e e n}{8.8 \cdot {10}^{- 8} \text{M}}$