Step 1: Calculate the $\text{p}K_a$ of acetic acid ($\text{CH}_3\text{CO}_2\text{H}$). The acid dissociation constant ($K_a$) for $\text{CH}_3\text{CO}_2\text{H}$ is given as $1.76 \times 10^{-5}$. $$ \text{p}K_a = -\log_{10}(K_a) $$ $$ \text{p}K_a = -\log_{10}(1.76 \times 10^{-5}) $$ $$ \text{p}K_a \approx 4.754 $$ Step 2: Use the Henderson-Hasselbalch equation to calculate the pH. 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 $5$. The final answer is $\boxed{5.01}$. Got more? Send 'em!