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

Previous problemNext problem

Struggling with General Chemistry?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

A 20.00-mL sample of 0.150 M NH3 is being titrated with 0.200 M HCl. What is the pH after 25.00 mL of HCl has been added? Kb of NH3 = 1.8 × 10⁻⁵

5.25

3.50

9.26

4.74

Previous problemNext problem

Verified step by step guidance

Calculate the initial moles of NH3 in the solution using the formula: \( \text{moles} = \text{volume} \times \text{molarity} \). For NH3, \( \text{moles} = 20.00 \text{ mL} \times 0.150 \text{ M} \). Convert the volume from mL to L by dividing by 1000.

Calculate the moles of HCl added using the same formula: \( \text{moles} = \text{volume} \times \text{molarity} \). For HCl, \( \text{moles} = 25.00 \text{ mL} \times 0.200 \text{ M} \). Again, convert the volume from mL to L.

Determine the limiting reactant by comparing the moles of NH3 and HCl. Since HCl is a strong acid and NH3 is a weak base, they will react in a 1:1 ratio to form NH4+. Subtract the moles of HCl from the moles of NH3 to find the remaining moles of NH3 or the moles of NH4+ formed.

Calculate the concentration of NH4+ in the solution. The total volume of the solution after the addition of HCl is the sum of the initial NH3 solution volume and the added HCl volume. Use this total volume to find the concentration: \( \text{concentration} = \frac{\text{moles of NH4+}}{\text{total volume in L}} \).

Use the Henderson-Hasselbalch equation to find the pH of the solution: \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \). First, calculate \( \text{pKa} \) from \( \text{pKb} \) using the relation \( \text{pKa} + \text{pKb} = 14 \). Then, substitute the concentrations of NH3 and NH4+ into the equation to find the pH.