Hypochlorous acid (HOCl) is a weak acid with pKa = 7.54. Suppose a 45.00 mL sample of 0.240 M HOCl solution is titrated with a 0.250 M standard KOH solution. What is the pH at the equivalence point?
The correct answer and explanation is :
At the equivalence point of a titration between a weak acid like hypochlorous acid (HOCl) and a strong base like potassium hydroxide (KOH), the resulting solution contains the conjugate base of the weak acid—in this case, the hypochlorite ion (OCl⁻). This conjugate base undergoes hydrolysis, reacting with water to produce hydroxide ions (OH⁻), which makes the solution basic. Consequently, the pH at the equivalence point will be greater than 7.
Calculating the pH at the Equivalence Point:
- Determine the Moles of HOCl:
- Volume of HOCl solution = 45.00 mL = 0.04500 L
- Molarity of HOCl = 0.240 M
- Moles of HOCl = Volume × Molarity = 0.04500 L × 0.240 mol/L = 0.0108 mol
- Calculate the Volume of KOH Needed:
- Molarity of KOH = 0.250 M
- Moles of KOH required = Moles of HOCl = 0.0108 mol
- Volume of KOH (L) = Moles / Molarity = 0.0108 mol / 0.250 mol/L = 0.0432 L
- Volume of KOH (mL) = 0.0432 L × 1000 mL/L = 43.2 mL
- Total Volume at Equivalence Point:
- Total volume = Volume of HOCl + Volume of KOH = 45.00 mL + 43.2 mL = 88.2 mL = 0.0882 L
- Concentration of OCl⁻ at Equivalence:
- Concentration of OCl⁻ = Moles of OCl⁻ / Total volume = 0.0108 mol / 0.0882 L ≈ 0.1225 M
- Calculate the pH:
- The hypochlorite ion (OCl⁻) is a weak base and will hydrolyze according to:
[ \text{OCl}^- + \text{H}_2\text{O} \rightleftharpoons \text{HOCl} + \text{OH}^- ] - The base dissociation constant (Kb) for OCl⁻ is related to the acid dissociation constant (Ka) of HOCl by:
[ K_b = \frac{K_w}{K_a} ]
Given that the pKa of HOCl is 7.54,
[ K_a = 10^{-7.54} ≈ 3.61 \times 10^{-8} ]
Therefore,
[ K_b = \frac{1.0 \times 10^{-14}}{3.61 \times 10^{-8}} ≈ 2.77 \times 10^{-7} ] - Setting up the equilibrium expression:
[ K_b = \frac{[\text{HOCl}][\text{OH}^-]}{[\text{OCl}^-]} ]
Assuming x is the concentration of OH⁻ produced:
[ 2.77 \times 10^{-7} = \frac{x^2}{0.1225} ]
Solving for x:
[ x^2 = 2.77 \times 10^{-7} \times 0.1225 = 3.40 \times 10^{-8} ]
[ x = \sqrt{3.40 \times 10^{-8}} ≈ 5.83 \times 10^{-4} \, \text{M} ] - The pOH is:
[ \text{pOH} = -\log(5.83 \times 10^{-4}) ≈ 3.23 ] - Finally, the pH is:
[ \text{pH} = 14 – \text{pOH} = 14 – 3.23 = 10.77 ]
Therefore, the pH at the equivalence point is approximately 10.77, indicating a basic solution due to the hydrolysis of the hypochlorite ion.