If 17.0 mL of glacial acetic acid (pure HC2H3O2; Ka=1.8×10⁻⁵) is diluted to 1.65 L with water, what is the pH of the resulting solution? The density of glacial acetic acid is 1.05 g/mL.

• A. 3.45
• B. 2.87
• C. 4.76
• D. 5.23

Answer: (B)

Verified step by step guidance

1. Calculate the mass of glacial acetic acid using its volume and density. Use the formula: \( \text{mass} = \text{volume} \times \text{density} \).
2. Convert the mass of acetic acid to moles using its molar mass. The molar mass of acetic acid (HC2H3O2) is approximately 60.05 g/mol. Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \).
3. Determine the concentration of acetic acid in the solution by dividing the number of moles by the total volume of the solution in liters. Use the formula: \( \text{concentration} = \frac{\text{moles}}{\text{volume}} \).
4. Use the acid dissociation constant (Ka) to set up an expression for the dissociation of acetic acid: \( \text{HC}_2\text{H}_3\text{O}_2 \rightleftharpoons \text{H}^+ + \text{C}_2\text{H}_3\text{O}_2^- \). Write the expression for Ka: \( K_a = \frac{[\text{H}^+][\text{C}_2\text{H}_3\text{O}_2^-]}{[\text{HC}_2\text{H}_3\text{O}_2]} \).
5. Assume \([\text{H}^+] = [\text{C}_2\text{H}_3\text{O}_2^-] = x\) and \([\text{HC}_2\text{H}_3\text{O}_2] = \text{initial concentration} - x\). Solve for \(x\) using the quadratic formula or by assuming \(x\) is small compared to the initial concentration. Calculate the pH using \( \text{pH} = -\log[\text{H}^+] \).