Table of contents

1. Matter and Measurements4h 31m
2. Atoms and the Periodic Table5h 23m
3. Ionic Compounds2h 18m
4. Molecular Compounds2h 18m
5. Classification & Balancing of Chemical Reactions3h 17m
6. Chemical Reactions & Quantities2h 27m
7. Energy, Rate and Equilibrium3h 46m
8. Gases, Liquids and Solids3h 25m
9. Solutions4h 10m
10. Acids and Bases3h 29m
11. Nuclear Chemistry56m
BONUS: Lab Techniques and Procedures1h 38m
BONUS: Mathematical Operations and Functions47m
12. Introduction to Organic Chemistry1h 34m
13. Alkenes, Alkynes, and Aromatic Compounds2h 12m
14. Compounds with Oxygen or Sulfur1h 6m
15. Aldehydes and Ketones1h 1m
16. Carboxylic Acids and Their Derivatives1h 11m
17. Amines39m
18. Amino Acids and Proteins1h 51m
19. Enzymes1h 37m
20. Carbohydrates1h 41m
21. The Generation of Biochemical Energy2h 8m
22. Carbohydrate Metabolism2h 22m
23. Lipids2h 26m
24. Lipid Metabolism1h 45m
25. Protein and Amino Acid Metabolism1h 37m
26. Nucleic Acids and Protein Synthesis2h 54m

10. Acids and Bases

The pH Scale

GOB Chemistry

10. Acids and Bases

The pH Scale

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1:24 minutes

Problem 34a

Textbook Question

Calculate the [H₃O⁺] of each aqueous solution with the following [OH^-]:
a. baking soda, 1.0 × 10^-6 M

Verified step by step guidance

Step 1: Recall the relationship between [H₃O⁺] and [OH⁻] in aqueous solutions, which is given by the ion-product constant for water:

K_w = [H₃O⁺][OH⁻]

. At 25°C,

K_w = 1.0 \times 10^{-14}

Step 2: Rearrange the equation to solve for [H₃O⁺]:

[H₃O⁺] = \frac{K_w}{[OH⁻]}

Step 3: Substitute the given value of [OH⁻] into the equation. For baking soda, [OH⁻] =

1.0 \times 10^{-6} \text{ M}

. The equation becomes:

[H₃O⁺] = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-6}}

Step 4: Simplify the expression by dividing the values in the numerator and denominator. Use the rules of exponents:

\frac{10^{-14}}{10^{-6}} = 10^{-14 + 6} = 10^{-8}

Step 5: Conclude that the [H₃O⁺] for the solution is

1.0 \times 10^{-8} \text{ M}

. This indicates the solution is slightly basic, as [H₃O⁺] is less than

1.0 \times 10^{-7} \text{ M}

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Key Concepts

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

pH and pOH Relationship

The pH and pOH of a solution are related through the equation pH + pOH = 14 at 25°C. This relationship allows us to calculate one from the other. Since pH measures the concentration of hydrogen ions [H₃O⁺] and pOH measures the concentration of hydroxide ions [OH⁻], knowing one enables the determination of the other, which is essential for understanding acid-base chemistry.

Ion Product of Water (Kw)

The ion product of water (Kw) is a constant that equals 1.0 x 10⁻¹⁴ at 25°C, representing the product of the concentrations of hydrogen ions [H₃O⁺] and hydroxide ions [OH⁻]. This means that in any aqueous solution, the product of [H₃O⁺] and [OH⁻] will always equal Kw. This concept is crucial for calculating the concentration of [H₃O⁺] when [OH⁻] is known.

Calculating [H₃O⁺] from [OH⁻]

To find the concentration of hydronium ions [H₃O⁺] from a given concentration of hydroxide ions [OH⁻], you can use the formula [H₃O⁺] = Kw / [OH⁻]. By substituting the known value of [OH⁻] into this equation, you can determine the corresponding [H₃O⁺] concentration, which is essential for understanding the acidity of the solution.

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