Multiple Choice
Calculate the concentration of the lactate ion in a solution that is 0.100 M in lactic acid (CH₃CH(OH)COOH, pKa = 3.86) and 0.080 M in HCl.
17. Acid and Base Equilibrium
pH of Weak Acids
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Calculate the pH of a 0.0095 M solution of acetic acid, CH3COOH, given that the acid dissociation constant (Ka) is 1.8 x 10^-5.
A
pH = 5.25
B
pH = 3.02
C
pH = 4.75
D
pH = 2.74
Verified step by step guidance1
Start by writing the chemical equation for the dissociation of acetic acid in water: \( \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \).
Set up the expression for the acid dissociation constant \( K_a \) using the formula: \( K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \).
Assume that the initial concentration of acetic acid is 0.0095 M and that the change in concentration due to dissociation is \( x \). Therefore, at equilibrium, \([\text{CH}_3\text{COO}^-] = x\), \([\text{H}^+] = x\), and \([\text{CH}_3\text{COOH}] = 0.0095 - x\).
Substitute these equilibrium concentrations into the \( K_a \) expression: \( 1.8 \times 10^{-5} = \frac{x^2}{0.0095 - x} \).
Assume \( x \) is small compared to 0.0095, so \( 0.0095 - x \approx 0.0095 \). Solve the simplified equation \( 1.8 \times 10^{-5} = \frac{x^2}{0.0095} \) for \( x \), and then calculate the pH using \( \text{pH} = -\log_{10}(x) \).
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