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

17. Acid and Base Equilibrium

pH of Weak Acids

General Chemistry

17. Acid and Base Equilibrium

pH of Weak Acids

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Multiple Choice

What is the pH of a 0.090 M H2CO3 solution, given that the first dissociation constant (Ka1) is 4.3×10^-7 and the second dissociation constant (Ka2) is 5.6×10^-11?

pH = 6.10

pH = 3.68

pH = 5.23

pH = 4.52

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Verified step by step guidance

Start by understanding that carbonic acid (H2CO3) is a diprotic acid, meaning it can donate two protons (H+). The dissociation occurs in two steps, each with its own dissociation constant: Ka1 for the first dissociation and Ka2 for the second.

For the first dissociation, write the equilibrium expression: H2CO3 ⇌ HCO3^- + H+. The equilibrium constant expression is given by:

K_{1} = \frac{[HCO3^-] \times [H+]}{[H2CO3]}

Assume that the initial concentration of H2CO3 is 0.090 M and that the change in concentration due to dissociation is small. Let x be the concentration of H+ ions produced from the first dissociation. Therefore, [HCO3^-] = x and [H+] = x, while [H2CO3] ≈ 0.090 M.

Substitute these values into the equilibrium expression for Ka1:

K_{1} = \frac{x \times x}{0.090} = 4.3 \times 10^{- 7}

. Solve for x, which represents [H+].

Calculate the pH using the formula:

pH = - \log ([H+])

. Since the second dissociation constant (Ka2) is much smaller, its contribution to the [H+] is negligible, and the pH is primarily determined by the first dissociation.