Questions: What is the approximate pH of a 0.10 M solution of a weak acid that has a Ka of 5 times 10^-5 M ?

Transcript text: What is the approximate pH of a 0.10 M solution of a weak acid that has a $\mathrm{K}_{\mathrm{a}}$ of $5 \times 10^{-5} \mathrm{M}$ ?

Solution

Solution Steps

Step 1: Understand the Problem

We need to find the pH of a 0.10 M solution of a weak acid with a given acid dissociation constant, $ K_a = 5 \times 10^{-5} \, \text{M} $.

Step 2: Set Up the Equilibrium Expression

For a weak acid $ HA $ dissociating in water: \[ HA \rightleftharpoons H^+ + A^- \]

The equilibrium expression for the dissociation is: \[ K_a = \frac{[H^+][A^-]}{[HA]} \]

Step 3: Make Approximations and Solve for $[H^+]$

Assume that the initial concentration of the acid, $[HA]_0 = 0.10 \, \text{M}$, changes by a small amount $ x $ upon dissociation: \[ K_a = \frac{x^2}{0.10 - x} \]

Since $ K_a $ is small, assume $ x \ll 0.10 $, so $ 0.10 - x \approx 0.10 $: \[ 5 \times 10^{-5} = \frac{x^2}{0.10} \]

Solving for $ x $: \[ x^2 = 5 \times 10^{-6} \] \[ x = \sqrt{5 \times 10^{-6}} \] \[ x \approx 2.2361 \times 10^{-3} \]

Step 4: Calculate the pH

The concentration of $ H^+ $ ions is approximately $ x $, so: \[ \text{pH} = -\log_{10}(x) \] \[ \text{pH} = -\log_{10}(2.2361 \times 10^{-3}) \] \[ \text{pH} \approx 2.65 \]