Calculate the pKa of acetic acid (CH3CO₂H).

The acid dissociation constant ($K_a$) for $\text{CH}_3\text{CO}_2\text{H}$ is given as $1.76 \times 10^{-5}$.

$$\text{p}K_a = -\log_{10}(K_a)$$ $$\text{p}K_a = -\log_{10}(1.76 \times 10^{-5})$$ $$\text{p}K_a \approx 4.754$$

The Henderson-Hasselbalch equation is:

$$\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)$$

We are given:
$[\text{HA}] = [\text{CH}_3\text{CO}_2\text{H}] = 0.25 \text{ M}$
$[\text{A}^-] = [\text{CH}_3\text{CO}_2^-] = 0.45 \text{ M}$
$\text{p}K_a = 4.754$
Substitute these values into the equation:

$$\text{pH} = 4.754 + \log_{10}\left(\frac{0.45 \text{ M}}{0.25 \text{ M}}\right)$$ $$\text{pH} = 4.754 + \log_{10}(1.8)$$ $$\text{pH} = 4.754 + 0.255$$ $$\text{pH} = 5.009$$

Rounding to a reasonable number of significant figures (e.g., two decimal places for pH), the pH is 5.01. If rounded to the nearest whole number as shown in the provided solution, it is 5.