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

Intro to Buffers

General Chemistry

18. Aqueous Equilibrium

Intro to Buffers

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Multiple Choice

Calculate the pH value for a buffer solution prepared by adding 1.0 mL of a 0.10M acetic acid solution to 20 mL of a 0.10M sodium acetate solution. (Ka = 1.7 × 10⁻⁵)

4.74

3.74

5.74

6.74

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Verified step by step guidance

First, understand that a buffer solution consists of a weak acid and its conjugate base. In this case, acetic acid (CH₃COOH) is the weak acid, and sodium acetate (CH₃COONa) provides the conjugate base, acetate ion (CH₃COO⁻).

Use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). Here, \( \text{pK}_a \) is the negative logarithm of the acid dissociation constant \( K_a \).

Calculate \( \text{pK}_a \) using the given \( K_a \) value: \( \text{pK}_a = -\log(1.7 \times 10^{-5}) \).

Determine the concentrations of acetic acid \([\text{HA}]\) and acetate ion \([\text{A}^-]\) in the buffer solution. Since the volumes are additive, calculate the final concentrations after mixing: \([\text{HA}] = \frac{0.10 \text{ M} \times 1.0 \text{ mL}}{21.0 \text{ mL}}\) and \([\text{A}^-] = \frac{0.10 \text{ M} \times 20.0 \text{ mL}}{21.0 \text{ mL}}\).

Substitute the values of \( \text{pK}_a \), \([\text{A}^-]\), and \([\text{HA}]\) into the Henderson-Hasselbalch equation to find the pH: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \).