Chemistry question

Question

The p𝐾a of hypochlorous acid is 7.530 A 57.0 mL solution of 0.136 M sodium hypochlorite (NaOCl) is titrated with 0.261 M HCl

1. Calculate the pH of the solution after the addition of 10.0 mL of 0.261 M HCl.
2. Calculate the pH of the solution after the addition of 30.9 mL of 0.261 M HCl.
3. Calculate the pH of the solution at the equivalence point with 0.261M HCl.

Answer 1

Addition of HCl (or simply H⁺) to ClO^- will result in formation of the weak acid, HClO:

Initial moles ClO- = 57.0 ml x 1 L / 1000 ml x 0.136 mol/L = 0.007752 mols ClO^-

H⁺ + ClO^- ==> HClO

1). 10.0 ml x 1 L / 1000 ml x 0.261 mol / L = 0.00261 moles H⁺

0.00775 mols ClO^- - 0.00261 mols = 0.00514 mol ClO^- + 0.00261 mol HClO in a volume of 57 + 10 = 67 ml

[ClO^-] = 0.00514 mol/0.067 L = 0.0767 M

[HClO] = 0.00261 mol/0.067 L = 0.0390 M

pH = pKa + log [0.0767]/[0.0390] ... Henderson Hasselbalch equation

pH = 7.53 + 0.29

pH = 7.82

2). 30.9 ml x 1 L / 1000 ml x 0.261 mol/L = 0.00806 mols H⁺

0.00806 mol H⁺ reacting with 0.00775 mols ClO^- => 0.00775 mol HClO + 0.00031 mol H⁺ in excess

Final volume = 57 ml + 30.9 ml = 87.9 ml

[H+] = 0.00031 mol / 0.0879 L = 0.00353 M

pH = -log 0.00353

pH = 2.45

3). At equivalence, moles H⁺ = moles ClO^- = 0.00775 mols

volume of HCl = 0.00775 mol x 1 L / 0.261 mol = 0.0297 L = 29.7 mls

At this point all H⁺ and ClO^- exist as HClO (the weak acid).

[HClO] = 0.00775 mols / 0.0297 L = 0.261 M

Ka = [H⁺][ClO^-] / [HClO] and Ka = 1x10^-7.53 = 2.95x10^-8

2.95x10^-8 = (x)(x) / 0.261-x (assume x is small relative to 0.261 and ignore it in denominator)

2.95x10^-8 = (x)(x) / 0.261

x² = 7.70x10^-9

x = [H+] = 8.77x10^-5 M

pH = -log 8.77x10^-5

pH = 4.06