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

A 100.0 mL sample of 0.20 mol L⁻¹ HF is titrated with 0.10 mol L⁻¹ KOH. Determine the pH of the solution after the addition of 200.0 mL of KOH. The Kₐ of HF is 3.5 × 10⁻⁴.

pH = 7.00

pH = 8.72

pH = 3.14

pH = 5.28

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Verified step by step guidance

Calculate the initial moles of HF using the formula: \( \text{moles of HF} = \text{concentration of HF} \times \text{volume of HF} \). Convert the volume from mL to L before calculation.

Calculate the moles of KOH added using the formula: \( \text{moles of KOH} = \text{concentration of KOH} \times \text{volume of KOH} \). Again, convert the volume from mL to L before calculation.

Determine the limiting reactant by comparing the initial moles of HF and the moles of KOH added. Since HF is a weak acid and KOH is a strong base, they will react in a 1:1 ratio to form water and the conjugate base, F⁻.

Calculate the moles of HF remaining and the moles of F⁻ formed after the reaction. If KOH is the limiting reactant, all of it will react, and the remaining moles of HF will be \( \text{initial moles of HF} - \text{moles of KOH} \). The moles of F⁻ formed will be equal to the moles of KOH added.

Use the Henderson-Hasselbalch equation to find the pH of the solution: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{F}^-]}{[\text{HF}]} \right) \). Calculate \( \text{pK}_a \) from \( K_a \) using \( \text{pK}_a = -\log(K_a) \). Substitute the concentrations of F⁻ and HF, which are the moles divided by the total volume of the solution (sum of the initial HF and added KOH volumes in L).