Questions: What is the pH of a 0.024 M acetic acid (Ka=1.8 x 10^-5) solution at 25°C ?

Transcript text: 17. What is the pH of a 0.024 M acetic acid $\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)$ solution at $25^{\circ} \mathrm{C}$ ? A: 3.18

Solution

Solution Steps

Step 1: Write the expression for the dissociation of acetic acid

Acetic acid ($\mathrm{CH_3COOH}$) dissociates in water as follows: \[ \mathrm{CH_3COOH} \rightleftharpoons \mathrm{CH_3COO^-} + \mathrm{H^+} \]

Step 2: Write the expression for the acid dissociation constant

The acid dissociation constant ($K_a$) for acetic acid is given by: \[ K_a = \frac{[\mathrm{CH_3COO^-}][\mathrm{H^+}]}{[\mathrm{CH_3COOH}]} \]

Step 3: Set up the equilibrium concentrations

Let $x$ be the concentration of $\mathrm{H^+}$ ions at equilibrium. Initially, the concentration of acetic acid is 0.024 M, and the concentrations of $\mathrm{CH_3COO^-}$ and $\mathrm{H^+}$ are 0. At equilibrium, the concentrations will be: \[ [\mathrm{CH_3COOH}] = 0.024 - x \] \[ [\mathrm{CH_3COO^-}] = x \] \[ [\mathrm{H^+}] = x \]

Step 4: Substitute the equilibrium concentrations into the $K_a$ expression

\[ K_a = \frac{x \cdot x}{0.024 - x} = 1.8 \times 10^{-5} \] \[ x^2 = 1.8 \times 10^{-5} \cdot (0.024 - x) \]

Step 5: Simplify the equation assuming $x$ is small

Since $K_a$ is small, we can assume $x$ is much smaller than 0.024, so $0.024 - x \approx 0.024$: \[ x^2 \approx 1.8 \times 10^{-5} \cdot 0.024 \] \[ x^2 \approx 4.32 \times 10^{-7} \] \[ x \approx \sqrt{4.32 \times 10^{-7}} \] \[ x \approx 6.57 \times 10^{-4} \]

Step 6: Calculate the pH

The pH is given by: \[ \text{pH} = -\log[\mathrm{H^+}] \] \[ \text{pH} = -\log(6.57 \times 10^{-4}) \] \[ \text{pH} \approx 3.18 \]