Calculate the \(\mathrm{pH}\) at the equivalence point in titrating \(0.100 \mathrm{M}\) solutions of each of the following with \(0.080 \mathrm{M} \mathrm{NaOH}\) : (a) hydrobromic acid (HBr), (b) chlorous acid (HClO \(_{2}\) ), (c) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\).

We will begin by determining whether each given acid is strong or weak. (a) Hydrobromic acid (HBr) is a strong acid. (b) Chlorous acid (HClO2) is a weak acid. (c) Benzoic acid (C6H5COOH) is a weak acid. Now that we have identified the acid strength, we can proceed with finding the pH at the equivalence point.

To calculate the moles of the acid and the base, we will use the formula: moles = concentration × volume For all three acids, let's assume that we started with a 50 mL (0.050 L) solution of 0.100 M acid. The moles of acid can be calculated as follows: moles (acid) = 0.100 M × 0.050 L = 0.005 mol Now, let's find the volume of NaOH needed to reach the equivalence point for each acid (at equivalence point, moles of acid = moles of base): moles (NaOH) = 0.005 mol Volume (NaOH) = moles (NaOH) / concentration (NaOH) = 0.005 mol / 0.080 M ≈ 0.0625 L

(a) For HBr, since it's a strong acid, it will completely dissociate: HBr + OH⁻ → Br⁻ + H₂O At equivalence point: [HBr] = 0 [OH⁻] = 0.005 mol / (0.050 L + 0.0625 L) = 0.044 M pOH = -log(0.044) pH = 14 - pOH ≈ 12.36 (b) For HClO2, the reaction with NaOH will produce its conjugate base ClO2⁻: HClO2 + OH⁻ → ClO2⁻ + H2O At equivalence point, all of the HClO2 will be converted to ClO2⁻: [ClO2⁻] = 0.005 mol / (0.050 L + 0.0625 L) = 0.044 M Now we have to use the Ka expression for HClO2 to calculate the [H+] concentration: Ka = [H+][ClO2⁻]/[HClO2] ≈ 1.1 × 10⁻² [H+] = sqrt(Ka × [ClO2⁻]) ≈ 6.371 × 10⁻² M pH = -log(6.371 × 10⁻²) ≈ 1.20 (c) For C6H5COOH, the reaction with NaOH will produce its conjugate base C6H5COO⁻: C6H5COOH + OH⁻ → C6H5COO⁻ + H₂O At equivalence point, all of the C6H5COOH will be converted to C6H5COO⁻: [C6H5COO⁻] = 0.005 mol / (0.050 L + 0.0625 L) = 0.044 M Now we have to use the Ka expression for C6H5COOH to calculate the [H+] concentration: Ka = [H+][C6H5COO⁻]/[C6H5COOH] ≈ 6.5 × 10⁻⁵ [H+] = sqrt(Ka × [C6H5COO⁻]) ≈ 1.034 × 10⁻² M pH = -log(1.034 × 10⁻²) ≈ 2.03 So, the pH at the equivalence point for each of the acids is: (a) Hydrobromic acid (HBr): pH ≈ 12.36 (b) Chlorous acid (HClO2): pH ≈ 1.20 (c) Benzoic acid (C6H5COOH): pH ≈ 2.03