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

A 1.00 L buffer solution is 0.250 M in HF and 0.250 M in NaF. Calculate the pH of the solution after the addition of 100.0 mL of 1.00 M HCl. The Ka for HF is 3.5 × 10⁻⁴.

2.92

3.46

3.74

3.14

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Verified step by step guidance

Identify the components of the buffer solution: HF (weak acid) and NaF (its conjugate base). The initial concentrations are both 0.250 M in a 1.00 L solution.

Calculate the moles of HF and NaF initially present in the buffer. Since the volume is 1.00 L, the moles of each are: \( \text{moles of HF} = 0.250 \text{ M} \times 1.00 \text{ L} = 0.250 \text{ moles} \) and \( \text{moles of NaF} = 0.250 \text{ M} \times 1.00 \text{ L} = 0.250 \text{ moles} \).

Determine the moles of HCl added: \( \text{moles of HCl} = 1.00 \text{ M} \times 0.100 \text{ L} = 0.100 \text{ moles} \). HCl is a strong acid and will react completely with the NaF (conjugate base).

Calculate the new moles of HF and NaF after the reaction with HCl. The reaction is: \( \text{F}^- + \text{HCl} \rightarrow \text{HF} + \text{Cl}^- \). Subtract the moles of HCl from NaF and add to HF: \( \text{moles of NaF} = 0.250 - 0.100 = 0.150 \text{ moles} \) and \( \text{moles of HF} = 0.250 + 0.100 = 0.350 \text{ moles} \).

Use the Henderson-Hasselbalch equation to find the pH: \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). Calculate \( \text{pKa} = -\log(3.5 \times 10^{-4}) \) and substitute the concentrations \( [\text{A}^-] = \frac{0.150}{1.10} \) and \( [\text{HA}] = \frac{0.350}{1.10} \) into the equation to find the pH.