# Question 58e5e

Jun 8, 2017

$3.5 \times {10}^{-} 8$

#### Explanation:

We're asked to find the acid-dissociation constant ${K}_{a}$ of hypochlorous acid, $\text{HOCl}$ with a given molar concentration and $\text{pH}$.

Let's first write the equilibrium reaction equation:

${\text{HOCl" (aq) rightleftharpoons "H"^+ (aq) + "ClO}}^{-} \left(a q\right)$

From this, we can write the equilibrium constant expression:

K_a = (["H"^+]["ClO"^(-)])/(["HOCl"]

From the given $\text{pH}$ we can calculate the hydrogen-ion concentration, $\left[{\text{H}}^{+}\right]$:

$\left[\text{H"^+] = 10^(-"pH}\right) = {10}^{- 3.9} = 1.26 \times {10}^{-} 4 M$

Let's create a makeshift I.C.E. chart using bullet points to illustrate the equilibrium changes:

Initial

• $\text{HOCl}$: $0.45 M$

• ${\text{H}}^{+}$: $0 M$

• $\text{ClO}$: $0 M$

Change

• $\text{HOCl}$: $- 1.26 \times {10}^{-} 4 M$

• ${\text{H}}^{+}$: $+ 1.26 \times {10}^{-} 4 M$

• $\text{ClO}$: $+ 1.26 \times {10}^{-} 4 M$

Final

• $\text{HOCl}$: $0.45 M - 1.26 \times {10}^{-} 4 M$

• ${\text{H}}^{+}$: $1.26 \times {10}^{-} 4 M$

• $\text{ClO}$: $1.26 \times {10}^{-} 4 M$

From these equilibrium concentrations, we can plug these in to the equilibrium expression to calculate the ${K}_{a}$:

K_a = ([1.26xx10^-4M][1.26xx10^-4M])/([0.45M-1.26xx10^-4M]) = color(red)(3.5xx10^-8#