Multiple Choice
Calculate the pH at 25°C of a solution prepared by adding 50.0 mL of 1.60 M NaOH solution to 0.500 L of 0.0850 M HC5H3O3. Furoic acid (HC5H3O3) has a Ka value of 6.76×10⁻⁴ at 25°C.
17. Acid and Base Equilibrium
pH of Weak Acids
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Calculate the pH of a 0.03 M aqueous solution of a weak acid, HA, that has a pKa of 3.39.
A
pH = 4.13
B
pH = 2.87
C
pH = 1.52
D
pH = 3.39
Verified step by step guidance1
Identify the relationship between pKa and Ka using the formula: \( \text{pKa} = -\log_{10}(K_a) \). Rearrange this to find \( K_a \) as \( K_a = 10^{-\text{pKa}} \).
Substitute the given pKa value into the equation to calculate \( K_a \): \( K_a = 10^{-3.39} \).
Set up the expression for the dissociation of the weak acid HA in water: \( \text{HA} \rightleftharpoons \text{H}^+ + \text{A}^- \).
Use the expression for the equilibrium constant \( K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \) and assume that \([\text{H}^+] = [\text{A}^-] = x\) and \([\text{HA}] = 0.03 - x\).
Solve the equation \( K_a = \frac{x^2}{0.03 - x} \) for \( x \), which represents the concentration of \( \text{H}^+ \), and then calculate the pH using \( \text{pH} = -\log_{10}(x) \).
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