Multiple Choice
What is the calculated pH after adding 2.5 mL of 0.5 M NaOH to 50.0 mL of 0.25 M acetic acid, assuming complete neutralization of the added base?
17. Acid and Base Equilibrium
pH of Weak Acids
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What is the pH of a 0.150 M solution of CH3COOH, given that the Ka of CH3COOH is 1.8 × 10⁻⁵?
A
3.45
B
1.50
C
2.87
D
4.74
Verified step by step guidance1
Identify the given values: the concentration of acetic acid (CH3COOH) is 0.150 M and the acid dissociation constant (Ka) is 1.8 × 10⁻⁵.
Write the balanced chemical equation for the dissociation of acetic acid in water: CH3COOH ⇌ CH3COO⁻ + H⁺.
Set up an ICE (Initial, Change, Equilibrium) table to determine the concentrations of the species at equilibrium. Initially, [CH3COOH] = 0.150 M, [CH3COO⁻] = 0, and [H⁺] = 0. As the reaction proceeds, let x be the change in concentration for CH3COO⁻ and H⁺, so at equilibrium, [CH3COOH] = 0.150 - x, [CH3COO⁻] = x, and [H⁺] = x.
Apply the expression for the acid dissociation constant: Ka = [CH3COO⁻][H⁺] / [CH3COOH]. Substitute the equilibrium concentrations into this expression: 1.8 × 10⁻⁵ = (x)(x) / (0.150 - x).
Assume x is small compared to 0.150, so 0.150 - x ≈ 0.150. This simplifies the equation to 1.8 × 10⁻⁵ = x² / 0.150. Solve for x, which represents the [H⁺] concentration, and then calculate the pH using the formula pH = -log[H⁺].
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