Multiple Choice
Consider the titration of 25.00 mL of 0.150 M hydrazoic acid (Ka = 1.90 x 10^-5) with 0.200 M NaOH. What is the pH of the solution at the equivalence point of the titration?
18. Aqueous Equilibrium
Titrations: Weak Acid-Strong Base
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For a titration of 1.0 M benzoic acid (C6H5COOH, Ka = 6.5×10^-5) with 1.0 M NaOH, what is the pH at the equivalence point?
A
5.28
B
9.25
C
8.72
D
7.00
Verified step by step guidance1
Identify the chemical reaction occurring during the titration: Benzoic acid (C6H5COOH) reacts with NaOH to form sodium benzoate (C6H5COONa) and water. The balanced equation is: C6H5COOH + NaOH → C6H5COONa + H2O.
Recognize that at the equivalence point, all the benzoic acid has been converted to its conjugate base, sodium benzoate. Therefore, the solution contains only the conjugate base, C6H5COO⁻.
Understand that the pH at the equivalence point is determined by the hydrolysis of the conjugate base, C6H5COO⁻. The hydrolysis reaction is: C6H5COO⁻ + H2O ⇌ C6H5COOH + OH⁻.
Calculate the Kb for the conjugate base using the relation: Kb = Kw / Ka, where Kw is the ion-product constant of water (1.0 × 10⁻¹⁴) and Ka is the acid dissociation constant for benzoic acid (6.5 × 10⁻⁵).
Set up the expression for the equilibrium concentration of OH⁻ using the Kb value and the initial concentration of the conjugate base. Use the formula: Kb = [OH⁻][C6H5COOH] / [C6H5COO⁻]. Solve for [OH⁻] and then calculate the pOH. Finally, convert pOH to pH using the relation: pH = 14 - pOH.
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