The pKa of hypochlorous acid is 7.530. A 53.0 mL solution of 0.100 M sodium hypochlorite (NaOCl) is titrated with 0.303 M HCl

The pKa of hypochlorous acid is 7.530. A 53.0 mL solution of 0.100 M sodium hypochlorite (NaOCl) is titrated with 0.303 M HCl. Calculate the pH of the solution

a) after the addition of 6.47 mL of 0.303 M HCl.

b) after the addition of 18.4 mL of 0.303 M HCl.

c) at the equivalence point with 0.303 M HCl.

The correct answer and explanation is :

In this titration, sodium hypochlorite (NaOCl), the sodium salt of hypochlorous acid (HOCl), is neutralized by hydrochloric acid (HCl). The reaction between NaOCl and HCl forms HOCl and NaCl:

[ \text{NaOCl} + \text{HCl} \rightarrow \text{HOCl} + \text{NaCl} ]

The pKa of HOCl is given as 7.530. We can apply the Henderson-Hasselbalch equation to calculate the pH at various stages of the titration:

[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{OCl}^-]}{[\text{HOCl}]} \right) ]

a) After adding 6.47 mL of 0.303 M HCl:

1. Initial moles of NaOCl: [ \text{moles of NaOCl} = 0.100\, \text{M} \times 0.053\, \text{L} = 0.00530\, \text{mol} ]
2. Moles of HCl added: [ \text{moles of HCl} = 0.303\, \text{M} \times 0.00647\, \text{L} = 0.00196\, \text{mol} ]
3. Moles of NaOCl remaining: [ \text{remaining NaOCl} = 0.00530\, \text{mol} – 0.00196\, \text{mol} = 0.00334\, \text{mol} ]
4. Moles of HOCl formed: [ \text{HOCl formed} = 0.00196\, \text{mol} ]
5. Total volume of the solution: [ \text{total volume} = 53.0\, \text{mL} + 6.47\, \text{mL} = 59.47\, \text{mL} = 0.05947\, \text{L} ]
6. Concentrations of HOCl and OCl⁻:

• [ [\text{HOCl}] = \frac{0.00196\, \text{mol}}{0.05947\, \text{L}} = 0.0329\, \text{M} ]
• [ [\text{OCl}^-] = \frac{0.00334\, \text{mol}}{0.05947\, \text{L}} = 0.0561\, \text{M} ]

1. pH calculation: [ \text{pH} = 7.530 + \log \left( \frac{0.0561}{0.0329} \right) = 7.530 + \log (1.705) \approx 7.530 + 0.232 = 7.762 ]

b) After adding 18.4 mL of 0.303 M HCl:

1. Moles of HCl added: [ \text{moles of HCl} = 0.303\, \text{M} \times 0.0184\, \text{L} = 0.00558\, \text{mol} ]
2. Moles of NaOCl remaining: [ \text{remaining NaOCl} = 0.00530\, \text{mol} – 0.00558\, \text{mol} = -0.00028\, \text{mol} ] A negative value indicates that all NaOCl has reacted, and excess HCl remains.
3. Excess moles of HCl: [ \text{excess HCl} = 0.00558\, \text{mol} – 0.00530\, \text{mol} = 0.00028\, \text{mol} ]
4. Total volume of the solution: [ \text{total volume} = 53.0\, \text{mL} + 18.4\, \text{mL} = 71.4\, \text{mL} = 0.0714\, \text{L} ]
5. Concentration of excess HCl: [ [\text{HCl}] = \frac{0.00028\, \text{mol}}{0.0714\, \text{L}} = 0.00392\, \text{M} ]
6. pH calculation: [ \text{pH} = -\log (0.00392) \approx 2.407 ]

c) At the equivalence point with 0.303 M HCl:

1. Volume of HCl required to reach equivalence: [ \text{volume of HCl} = \frac{0.00530\, \text{mol}}{0.303\, \text{M}} = 0.0175\, \text{L} = 17.5\, \text{mL} ]
2. Concentration of HOCl at equivalence: [ [\text{HOCl}] = \frac{0.00530\, \text{mol}}{0.053\, \text{L} + 0.0175\, \text{L}} = \frac{0.00530\, \text{mol}}{0.0705\, \text{L}} \approx 0.0751\, \text{M} ]
3. pH calculation using the formula for a weak acid: [ \text{pH} = \frac{1}{2} \left( \text{pKa} – \log [\text{HOCl}] \right) = \frac{1}{2} \left( 7.530 – \log (0.0751)