Table of contents

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

3. Acids and Bases

Acids and Bases

Organic Chemistry

3. Acids and Bases

Acids and Bases

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7:57 minutes

Problem 8

Textbook Question

Calculate the pH of the following solutions.
a. 5.00 g of HBr in 100 mL of aqueous solution
b. 1.50 g of NaOH in 50 mL of aqueous solution

Verified step by step guidance

Step 1: Begin by calculating the molarity of the HBr solution. First, determine the number of moles of HBr using its molar mass. The molar mass of HBr is approximately 80.91 g/mol. Use the formula: \( \text{moles of HBr} = \frac{\text{mass of HBr}}{\text{molar mass of HBr}} \).

Step 2: Convert the volume of the solution from milliliters to liters. Since the volume is given as 100 mL, convert it to liters by dividing by 1000: \( \text{volume in liters} = \frac{100}{1000} \).

Step 3: Calculate the molarity of the HBr solution using the formula: \( \text{molarity} = \frac{\text{moles of HBr}}{\text{volume in liters}} \). Since HBr is a strong acid, it dissociates completely in water, and the concentration of \( \text{H}^+ \) ions will be equal to the molarity of the HBr solution.

Step 4: Calculate the pH of the HBr solution using the formula: \( \text{pH} = -\log[\text{H}^+] \), where \([\text{H}^+]\) is the molarity of the HBr solution.

Step 5: For the NaOH solution, repeat a similar process. Calculate the moles of NaOH using its molar mass (approximately 40.00 g/mol). Convert the volume from mL to liters (50 mL to 0.050 L). Calculate the molarity of NaOH, and since NaOH is a strong base, it dissociates completely, providing \([\text{OH}^-]\) equal to the molarity. Use the formula \( \text{pOH} = -\log[\text{OH}^-] \) and then find the pH using \( \text{pH} = 14 - \text{pOH} \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molarity

Molarity is a measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution (mol/L). To calculate molarity, you need to know the amount of solute in moles and the volume of the solution in liters. This concept is crucial for determining the concentration of HBr and NaOH in the given solutions.

pH Calculation

pH is a measure of the acidity or basicity of a solution, calculated as the negative logarithm of the hydrogen ion concentration (-log[H+]). For strong acids like HBr, which dissociate completely, the concentration of H+ ions is equal to the molarity of the acid. For strong bases like NaOH, the concentration of OH- ions can be used to find pH by first calculating pOH and then using the relation pH + pOH = 14.

Strong Acid and Base Dissociation

Strong acids and bases dissociate completely in aqueous solutions, meaning they fully ionize into their constituent ions. HBr, a strong acid, dissociates into H+ and Br- ions, while NaOH, a strong base, dissociates into Na+ and OH- ions. Understanding this complete dissociation is essential for accurately calculating the concentration of ions in the solution, which directly affects the pH.

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