Multiple Choice
Calculate the pH at 25°C of a 0.21 M solution of a weak acid that has Ka = 9.2 x 10⁻⁶.
17. Acid and Base Equilibrium
pH of Weak Acids
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What is the pH of a 0.10 M solution of hypochlorous acid (HOCl) given that the acid dissociation constant (Ka) is 3.0 x 10^-8?
A
5.00
B
2.00
C
4.52
D
7.00
Verified step by step guidance1
Start by writing the chemical equation for the dissociation of hypochlorous acid (HOCl) in water: \( \text{HOCl} \rightleftharpoons \text{H}^+ + \text{OCl}^- \).
Use the expression for the acid dissociation constant \( K_a \), which is given by \( K_a = \frac{[\text{H}^+][\text{OCl}^-]}{[\text{HOCl}]} \). Substitute the given \( K_a = 3.0 \times 10^{-8} \).
Assume that the initial concentration of HOCl is 0.10 M and that the change in concentration due to dissociation is \( x \). Therefore, at equilibrium, \([\text{H}^+] = x\), \([\text{OCl}^-] = x\), and \([\text{HOCl}] = 0.10 - x\).
Substitute these equilibrium concentrations into the \( K_a \) expression: \( 3.0 \times 10^{-8} = \frac{x^2}{0.10 - x} \). Since \( K_a \) is very small, assume \( x \ll 0.10 \), simplifying the expression to \( 3.0 \times 10^{-8} = \frac{x^2}{0.10} \).
Solve for \( x \) to find \([\text{H}^+]\), and then calculate the pH using the formula \( \text{pH} = -\log[\text{H}^+] \).
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