Question

The Ka of hypochlorous acid (HClO) is 3.00*10^-8. What is the pH at 25.0 °C of an aqueous solution that is 0.02M in HClO?​

Answer 1

Approximately
`4.6`.

Step-by-step explanation:

Hypochlorous acid
`\rm HClO` ionizes partially at room temperature:

`\rm HClO \rightleftharpoons H^(+) + ClO^(-)`.

The initial concentration of
`\rm HClO` in this solution is
`0.02\; \rm mol \cdot L^(-1)`.

Construct a
`\verb!RICE!` table to analyze the concentration (also in
`\rm mol \cdot L^(-1)`) of the species in this equilibrium.

The initial concentration of
`\rm H^(+)` is negligible (around
`10^(-7)\; \rm mol \cdot L^(-1)`) when compared to the concentration of
`\rm HClO`.

Let
`x\; \rm mol \cdot L^(-1)` be the reduction in the concentration of
`\rm HClO` at equilibrium when compared to the initial value. Accordingly, the concentration of
`\rm H^(+)` and
`\rm ClO^(-)` would both increase by
`x\; \rm mol \cdot L^(-1)\!`. (
`x > 0` since concentration should be non-negative.)

`\begin{array}ccccc\text{Reaction} & \rm HClO & \rightleftharpoons & \rm H^(+) & + & \rm ClO^(-) \\ \text{Initial} & 0.02 & & & &x \\ \text{Change} & -x & & +x & & +x \\ \text{Equilibrium} & 0.02 - x & & x & & x\end{array}`.

Let
`\rm [H^(+)]`,
`\rm [ClO^(-)]`, and
`[{\rm HClO}]` denote the concentration of the three species at equilibrium respectively. Equation for the
`K_\text{a}` of
`\rm HClO`:

`\begin{aligned}K_\text{a} &= (\rm [H^(+)] \cdot [ClO^(-)])/([\rm HClO])\end{aligned}`.

Using equilibrium concentration values from the
`\verb!RICE!` table above:

`\begin{aligned}K_\text{a} &= (\rm [H^(+)] \cdot [ClO^(-)])/([\rm HClO]) = (x^(2))/(0.02 - x)\end{aligned}`.

`\begin{aligned}(x^(2))/(0.02 - x) &= 3.00 * 10^(-8)\end{aligned}`.

Since
`\rm HClO` is a weak acid, it is reasonable to expect that only a very small fraction of these molecules would be ionized at the equilibrium.

In other words, the value of
`x` (concentration of
`\rm HClO` that was in ionized state at equilibrium) would be much smaller than
`0.02` (initial concentration.)

Hence, it would be reasonable to estimate
`(0.02 - x)` as
`0.02`:

`\begin{aligned}(x^(2))/(0.02) &\approx (x^(2))/(0.02 - x) = 3.00 * 10^(-8)\end{aligned}`.

Solve for
`x` with the simplifying assumption:

`\begin{aligned}x &\approx \sqrt{0.02 * {3.00 * 10^(-8))}}\\ &\approx 2.45 * 10^(-5)\end{aligned}`.

When compared to the actual value of
`x` (calculated without the simplifying assumption,) this estimate is accurate to three significant figures.

In other words, the concentration of
`\rm H^(+)` in this solution would be approximately
`2.45 * 10^(-5)\; \rm mol \cdot L^(-1)` at equilibrium.

Hence the
`\text{pH}`:

`\begin{aligned}\text{pH} &= \log_(10) ([{\rm H^(+)}]) \\ &\approx \log_(10) (2.45 * 10^(-5)) \\ &\approx 4.6\end{aligned}`.