Table of contents

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1. Chemical Measurements1h 50m
2. Tools of the Trade1h 17m
3. Experimental Error1h 52m
4 & 5. Statistics, Quality Assurance and Calibration Methods1h 57m
6. Chemical Equilibrium3h 41m
7. Activity and the Systematic Treatment of Equilibrium1h 0m
8. Monoprotic Acid-Base Equilibria1h 53m
9. Polyprotic Acid-Base Equilibria2h 17m
10. Acid-Base Titrations2h 37m
11. EDTA Titrations1h 34m
12. Advanced Topics in Equilibrium1h 16m
13. Fundamentals of Electrochemistry2h 19m
14. Electrodes and Potentiometry41m
15. Redox Titrations1h 14m
16. Electroanalytical Techniques57m
17. Fundamentals of Spectrophotometry50m

8. Monoprotic Acid-Base Equilibria

Weak Acid-Base Equilibria

Analytical Chemistry

8. Monoprotic Acid-Base Equilibria

Weak Acid-Base Equilibria

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Multiple Choice

The pH of an aqueous 0.10 M nitrite ion is 8.17. What is the base dissociation constant of the base?

4.6 x 10^-16

2.2 x 10^-11

1.6 x 10^-6

1.6 x 10^‑5

1.2 x 10^-3

Verified step by step guidance

Identify the chemical species involved: The nitrite ion (NO2-) acts as a base in water, accepting a proton to form HNO2.

Use the given pH to find the concentration of hydroxide ions (OH-): Since pH + pOH = 14, calculate pOH as 14 - 8.17.

Convert pOH to [OH-]: Use the formula [OH-] = 10^(-pOH) to find the concentration of hydroxide ions.

Write the base dissociation equation: NO2- + H2O ⇌ HNO2 + OH-. The equilibrium expression for the base dissociation constant (Kb) is Kb = [HNO2][OH-] / [NO2-].

Substitute known values into the Kb expression: Use the initial concentration of NO2- (0.10 M) and the calculated [OH-] to solve for Kb, assuming [HNO2] ≈ [OH-] at equilibrium.