Multiple Choice
Calculate the pH of a solution that is 0.250 M in acetic acid (Ka = 1.8 × 10^-5) and 0.125 M in HClO (Ka = 2.9 × 10^-8).
17. Acid and Base Equilibrium
pH of Weak Acids
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What is the pH of a 0.1 M solution of acetic acid (CH3COOH) given that its acid dissociation constant (Ka) is 1.8 x 10^-5?
A
2.87
B
5.25
C
3.45
D
4.75
Verified step by step guidance1
Start by writing the chemical equation for the dissociation of acetic acid: \( \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \).
Use the expression for the acid dissociation constant \( K_a \), which is given by \( K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \). Substitute \( K_a = 1.8 \times 10^{-5} \) and the initial concentration of acetic acid \( [\text{CH}_3\text{COOH}] = 0.1 \text{ M} \).
Assume that the concentration of \( \text{CH}_3\text{COO}^- \) and \( \text{H}^+ \) ions formed is \( x \). Therefore, \( [\text{CH}_3\text{COO}^-] = x \) and \( [\text{H}^+] = x \), while \( [\text{CH}_3\text{COOH}] = 0.1 - x \).
Substitute these values into the \( K_a \) expression: \( 1.8 \times 10^{-5} = \frac{x^2}{0.1 - x} \). Since \( x \) is small compared to 0.1, you can approximate \( 0.1 - x \approx 0.1 \).
Solve for \( x \) to find \( [\text{H}^+] \), and then calculate the pH using the formula \( \text{pH} = -\log[\text{H}^+] \).
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