Question

Question #58e5e

Answer 1

`3.5xx10^-8`

Explanation:

We're asked to find the acid-dissociation constant `K_a` of hypochlorous acid, `"HOCl"` with a given molar concentration and `"pH"`.

Let's first write the equilibrium reaction equation:

`"HOCl" (aq) rightleftharpoons "H"^+ (aq) + "ClO"^(-) (aq)`

From this, we can write the equilibrium constant expression:

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

From the given `"pH"` we can calculate the hydrogen-ion concentration, `["H"^+]`:

`["H"^+] = 10^(-"pH") = 10^(-3.9) = 1.26 xx 10^-4 M`

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

Initial

• `"HOCl"`: `0.45M`

• `"H"^+`: `0M`

• `"ClO"`: `0M`

Change

• `"HOCl"`: `-1.26xx10^-4M`

• `"H"^+`: `+1.26 xx 10^-4M`

• `"ClO"`: `+1.26 xx 10^-4M`

Final

• `"HOCl"`: `0.45M - 1.26xx10^-4M`

• `"H"^+`: `1.26xx10^-4M`

• `"ClO"`: `1.26xx10^-4M`

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`