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 Base-Strong Acid

General Chemistry

18. Aqueous Equilibrium

Titrations: Weak Base-Strong Acid

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Multiple Choice

What is the pH of a solution when 40.00 mL of 0.1000 M HCl is added to a 50.00 mL aliquot of 0.1200 M ethylamine (CH3CH2NH2) with a Kb of 5.6 x 10^-4?

7.50

9.25

8.75

8.15

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Verified step by step guidance

Calculate the moles of HCl added using the formula: \( \text{moles of HCl} = \text{volume (L)} \times \text{molarity (M)} \). Convert 40.00 mL to liters and multiply by 0.1000 M.

Calculate the moles of ethylamine (CH3CH2NH2) using the formula: \( \text{moles of ethylamine} = \text{volume (L)} \times \text{molarity (M)} \). Convert 50.00 mL to liters and multiply by 0.1200 M.

Determine the limiting reactant by comparing the moles of HCl and ethylamine. The reaction is: \( \text{HCl} + \text{CH3CH2NH2} \rightarrow \text{CH3CH2NH3}^+ + \text{Cl}^- \). Subtract the smaller number of moles from both reactants to find the moles of excess reactant.

Calculate the concentration of the conjugate acid (CH3CH2NH3^+) formed. Use the volume of the final solution (sum of the initial volumes) to find the concentration: \( \text{concentration} = \frac{\text{moles of CH3CH2NH3}^+}{\text{total volume (L)}} \).

Use the Henderson-Hasselbalch equation to find the pH: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \). First, calculate \( \text{pK}_a \) from \( \text{pK}_a = 14 - \text{pK}_b \), where \( \text{pK}_b = -\log(5.6 \times 10^{-4}) \). Then, substitute the concentrations of the base (ethylamine) and the conjugate acid (CH3CH2NH3^+) into the equation.