Multiple Choice
Consider the titration of a 24.0 mL sample of 0.110 mol/L CH3COOH (Ka=1.8×10⁻⁵) with 0.125 mol/L NaOH. What volume of NaOH is required to reach the equivalence point?
344
views
18. Aqueous Equilibrium
Titrations: Weak Acid-Strong Base
Multiple Choice
Suppose 40.0 mL of an acetic acid solution of unknown concentration is titrated with 0.100 M NaOH. After 20.0 mL of the base solution has been added, the pH in the titration flask is 5.10. What was the concentration of the original acetic acid solution?
A
0.025 M
B
0.050 M
C
0.075 M
D
0.100 M
Verified step by step guidance1
Start by writing the balanced chemical equation for the reaction between acetic acid (CH₃COOH) and sodium hydroxide (NaOH): CH₃COOH + NaOH → CH₃COONa + H₂O.
Determine the moles of NaOH added using the formula: \( \text{moles of NaOH} = \text{volume (L)} \times \text{molarity (M)} \). Convert 20.0 mL to liters and multiply by 0.100 M.
Since the reaction is a 1:1 stoichiometry, the moles of NaOH added equals the moles of acetic acid that have reacted. Use this information to find the moles of acetic acid initially present.
Use the Henderson-Hasselbalch equation to relate the pH to the concentration of acetic acid and its conjugate base: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). Given \( \text{pH} = 5.10 \) and \( \text{pK}_a \approx 4.76 \) for acetic acid, solve for the ratio \( \frac{[\text{A}^-]}{[\text{HA}]} \).
Calculate the initial concentration of acetic acid using the moles of acetic acid and the initial volume of the solution (40.0 mL converted to liters). This will give you the concentration of the original acetic acid solution.
Related Videos
Related Practice
