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

6. Chemical Quantities & Aqueous Reactions

Dilutions

General Chemistry

6. Chemical Quantities & Aqueous Reactions

Dilutions

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Multiple Choice

Calculate the concentration (M) of sodium ions in a solution made by diluting 80.0 mL of a 0.174 M solution of sodium sulfide (Na2S) to a total volume of 125 mL.

0.111 M

0.087 M

0.070 M

0.139 M

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Verified step by step guidance

Start by identifying the initial concentration and volume of the sodium sulfide solution. The initial concentration is 0.174 M and the initial volume is 80.0 mL.

Calculate the number of moles of sodium sulfide (Na2S) in the initial solution using the formula: \( \text{moles} = \text{concentration} \times \text{volume} \). Convert the volume from mL to L by dividing by 1000.

Recognize that sodium sulfide (Na2S) dissociates into two sodium ions (Na\(^+\)) and one sulfide ion (S\(^{2-}\)) in solution. Therefore, the concentration of sodium ions will be twice the concentration of sodium sulfide.

Determine the concentration of sodium ions in the initial solution by multiplying the concentration of sodium sulfide by 2.

Calculate the final concentration of sodium ions after dilution using the formula: \( C_1V_1 = C_2V_2 \), where \( C_1 \) and \( V_1 \) are the initial concentration and volume, and \( C_2 \) and \( V_2 \) are the final concentration and volume. Solve for \( C_2 \) using the final volume of 125 mL (converted to L).