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1. Introduction to Biochemistry4h 34m
2. Water3h 23m
3. Amino Acids8h 10m
4. Protein Structure10h 4m
5. Protein Techniques14h 5m
6. Enzymes and Enzyme Kinetics13h 38m
7. Enzyme Inhibition and Regulation 8h 42m
8. Protein Function 9h 41m
9. Carbohydrates7h 49m
10. Lipids5h 49m
11. Biological Membranes and Transport 6h 37m
12. Biosignaling9h 45m
Review 1: Nucleic Acids, Lipids, & Membranes2h 47m
Review 2: Biosignaling, Glycolysis, Gluconeogenesis, & PP-Pathway3h 12m
Review 3: Pyruvate & Fatty Acid Oxidation, Citric Acid Cycle, & Glycogen Metabolism2h 26m
Review 4: Amino Acid Oxidation, Oxidative Phosphorylation, & Photophosphorylation1h 48m

2. Water

Titration

Biochemistry

2. Water

Titration

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Multiple Choice

Titration confirms an acetic acid solution to be 0.1 M. Calculate the pH. (acetic acid K _a = 1.76 x 10^-5 M).

2.1

3.6

2.9

8.3

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Verified step by step guidance

Identify the given values: the concentration of acetic acid (CH3COOH) is 0.1 M, and the acid dissociation constant (Ka) for acetic acid is 1.76 x 10^-5 M.

Write the equilibrium expression for the dissociation of acetic acid: CH3COOH ⇌ H+ + CH3COO-. The equilibrium constant expression is Ka = [H+][CH3COO-]/[CH3COOH].

Assume that the initial concentration of H+ and CH3COO- is 0, and let x be the concentration of H+ and CH3COO- at equilibrium. Therefore, [H+] = [CH3COO-] = x and [CH3COOH] = 0.1 - x.

Substitute these values into the equilibrium expression: 1.76 x 10^-5 = (x)(x)/(0.1 - x). Since Ka is small, assume x << 0.1, simplifying the expression to 1.76 x 10^-5 ≈ x^2/0.1.

Solve for x, which represents the concentration of H+ ions. Then, calculate the pH using the formula pH = -log[H+].