Multiple Choice
Calculate the pH of a 0.250 M CH3NH3I solution, given that the Kb for CH3NH2 is 4.4 x 10^-4.
17. Acid and Base Equilibrium
pH of Weak Acids
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What is the pH of a 0.040 M solution of benzoic acid, given that the acid dissociation constant (Ka) for benzoic acid is 6.3 x 10^-5?
A
5.00
B
2.52
C
4.18
D
3.45
Verified step by step guidance1
Start by writing the chemical equation for the dissociation of benzoic acid (C6H5COOH) in water: C6H5COOH ⇌ C6H5COO⁻ + H⁺.
Use the expression for the acid dissociation constant (Ka) to set up the equilibrium expression: Ka = [C6H5COO⁻][H⁺] / [C6H5COOH].
Assume that the initial concentration of benzoic acid is 0.040 M and that the change in concentration due to dissociation is 'x'. Therefore, at equilibrium, [C6H5COOH] = 0.040 - x, [C6H5COO⁻] = x, and [H⁺] = x.
Substitute these equilibrium concentrations into the Ka expression: 6.3 x 10^-5 = (x)(x) / (0.040 - x).
Solve the quadratic equation for 'x', which represents the concentration of H⁺ ions. Once 'x' is found, calculate the pH using the formula: pH = -log10([H⁺]).
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