Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 36m
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

Henderson-Hasselbalch Equation

General Chemistry

18. Aqueous Equilibrium

Henderson-Hasselbalch Equation

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 1.0 L buffer solution contains 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2. The value of Ka for HC2H3O2 is 1.8 × 10⁻⁵. Calculate the pH of the solution upon the addition of 0.015 mol of NaOH to the original buffer.

5.00

4.74

4.59

4.89

Previous problemNext problem

Verified step by step guidance

Identify the components of the buffer solution: acetic acid (HC2H3O2) and its conjugate base, sodium acetate (NaC2H3O2). The buffer solution initially contains 0.100 mol of each component in 1.0 L of solution.

Understand that the addition of NaOH will react with the acetic acid (HC2H3O2) in the buffer. The reaction is: \( \text{HC}_2\text{H}_3\text{O}_2 + \text{OH}^- \rightarrow \text{C}_2\text{H}_3\text{O}_2^- + \text{H}_2\text{O} \). This will decrease the amount of acetic acid and increase the amount of acetate ion.

Calculate the change in moles of acetic acid and acetate ion due to the addition of NaOH. Since 0.015 mol of NaOH is added, it will react with 0.015 mol of HC2H3O2, reducing its concentration to 0.085 mol and increasing the concentration of NaC2H3O2 to 0.115 mol.

Use the Henderson-Hasselbalch equation to calculate the pH of the buffer solution: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). Here, \( \text{pK}_a = -\log(1.8 \times 10^{-5}) \), \([\text{A}^-] = 0.115 \text{ mol/L}\), and \([\text{HA}] = 0.085 \text{ mol/L}\).

Substitute the values into the Henderson-Hasselbalch equation to find the pH: \( \text{pH} = \text{pK}_a + \log \left( \frac{0.115}{0.085} \right) \). Calculate \( \text{pK}_a \) and the logarithmic term to determine the pH.