Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

18. Aqueous Equilibrium

Titrations: Weak Acid-Strong Base

General Chemistry

18. Aqueous Equilibrium

Titrations: Weak Acid-Strong Base

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Multiple Choice

What is the pH at the equivalence point for a titration involving 25 mL of 0.1 M acetic acid and 0.2 M KOH, given that the acid dissociation constant (Ka) for acetic acid is 1.8 × 10^-5?

5.28

7.00

3.75

8.72

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Verified step by step guidance

Determine the moles of acetic acid initially present using the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). For acetic acid, \( \text{moles} = 0.1 \text{ M} \times 0.025 \text{ L} \).

Calculate the volume of KOH needed to reach the equivalence point. Since the reaction is a 1:1 molar ratio, the moles of KOH needed are equal to the moles of acetic acid. Use the formula: \( \text{volume} = \frac{\text{moles}}{\text{concentration}} \) for KOH.

At the equivalence point, all acetic acid is converted to acetate ion. The solution now contains the acetate ion, which is the conjugate base of acetic acid. Calculate the concentration of acetate ion by dividing the moles of acetate by the total volume of the solution at the equivalence point.

Use the hydrolysis of the acetate ion to find the \( \text{pH} \). The reaction is: \( \text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \). Use the expression for the base dissociation constant \( K_b \), which is related to \( K_a \) by \( K_w = K_a \times K_b \), where \( K_w = 1.0 \times 10^{-14} \).

Calculate the \( \text{pOH} \) from the concentration of \( \text{OH}^- \) ions using the formula \( \text{pOH} = -\log[\text{OH}^-] \), and then find the \( \text{pH} \) using \( \text{pH} = 14 - \text{pOH} \).