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

Answer:

Approximately $4.6$.

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.)

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

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 $\rm HClO$:

$\begin{aligned}K_\text{a} &= \frac{\rm [H^{+}] \cdot [ClO^{-}]}{[\rm HClO]}\end{aligned}$.

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

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

$\begin{aligned}\frac{x^{2}}{0.02 - x} &= 3.00 \times 10^{-8}\end{aligned}$.

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 $0.02$:

$\begin{aligned}\frac{x^{2}}{0.02} &\approx \frac{x^{2}}{0.02 - x} = 3.00 \times 10^{-8}\end{aligned}$.

Solve for $x$ with the simplifying assumption:

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

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 $2.45 \times 10^{-5}\; \rm mol \cdot L^{-1}$ at equilibrium.

Hence the $\text{pH}$:

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