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

17. Acid and Base Equilibrium

pH of Weak Bases

General Chemistry

17. Acid and Base Equilibrium

pH of Weak Bases

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Multiple Choice

Determine the pH of a solution made by dissolving 6.1 g of sodium cyanide, NaCN, in enough water to make a 500.0 mL of solution. (MW of NaCN = 49.01 g/mol). The K_a value of HCN is 4.9 × 10⁻¹⁰.

2.648

13.389

5.294

11.352

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Verified step by step guidance

Calculate the number of moles of NaCN by dividing the mass of NaCN (6.1 g) by its molar mass (49.01 g/mol). Use the formula: \( \text{moles of NaCN} = \frac{6.1 \text{ g}}{49.01 \text{ g/mol}} \).

Determine the concentration of NaCN in the solution by dividing the number of moles of NaCN by the volume of the solution in liters (0.500 L). Use the formula: \( \text{[NaCN]} = \frac{\text{moles of NaCN}}{0.500 \text{ L}} \).

Recognize that NaCN is a salt that dissociates completely in water to form Na⁺ and CN⁻ ions. The CN⁻ ion is the conjugate base of the weak acid HCN, and it will undergo hydrolysis in water to form OH⁻ ions, increasing the pH.

Use the hydrolysis equation for CN⁻: \( \text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^- \). The equilibrium expression for this reaction is \( K_b = \frac{[\text{HCN}][\text{OH}^-]}{[\text{CN}^-]} \). Calculate \( K_b \) using the relation \( K_w = K_a \times K_b \), where \( K_w = 1.0 \times 10^{-14} \) and \( K_a = 4.9 \times 10^{-10} \).

Set up an ICE table (Initial, Change, Equilibrium) for the hydrolysis reaction to find the concentration of OH⁻ at equilibrium. Use the expression for \( K_b \) to solve for \([\text{OH}^-]\), and then calculate the pOH of the solution. Finally, convert pOH to pH using the relation \( \text{pH} = 14 - \text{pOH} \).