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: Strong Acid-Strong Base

General Chemistry

18. Aqueous Equilibrium

Titrations: Strong Acid-Strong Base

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Multiple Choice

Two 20.0-mL samples, one 0.200 M KOH and the other 0.200 M unknown base, are titrated with 0.100 M HI. What is the volume of added acid at the equivalence point for each titration?

40.0 mL for both KOH and the unknown base

40.0 mL for KOH and 20.0 mL for the unknown base

20.0 mL for KOH and 40.0 mL for the unknown base

20.0 mL for both KOH and the unknown base

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Verified step by step guidance

Identify the reaction involved in the titration. For KOH and HI, the reaction is: \( \text{KOH} + \text{HI} \rightarrow \text{KI} + \text{H}_2\text{O} \). This is a neutralization reaction where one mole of KOH reacts with one mole of HI.

Calculate the moles of KOH present in the 20.0 mL sample. Use the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). Here, \( \text{concentration} = 0.200 \text{ M} \) and \( \text{volume} = 20.0 \text{ mL} = 0.0200 \text{ L} \).

Determine the volume of 0.100 M HI needed to reach the equivalence point for KOH. Since the reaction is 1:1, the moles of HI required will be equal to the moles of KOH. Use the formula: \( \text{volume of HI} = \frac{\text{moles of KOH}}{\text{concentration of HI}} \).

Repeat the calculation for the unknown base, assuming it also reacts in a 1:1 ratio with HI. Calculate the moles of the unknown base using its concentration and volume, then determine the volume of HI needed using the same method as for KOH.

Compare the calculated volumes of HI needed for both KOH and the unknown base to the given answer choices to determine the correct answer.