First, calculate the moles of hypochlorous acid (\( \text{HClO} \)) in the initial solution. Use the formula:
\[
\text{moles of HClO} = \text{volume (L)} \times \text{concentration (M)}
\]
Given:
- Volume of \(\text{HClO} = 29.8 \, \text{mL} = 0.0298 \, \text{L}\)
- Concentration of \(\text{HClO} = 0.483 \, \text{M}\)
\[
\text{moles of HClO} = 0.0298 \, \text{L} \times 0.483 \, \text{M} = 0.0143834 \, \text{mol}
\]
Next, calculate the moles of sodium hydroxide (\( \text{NaOH} \)) needed to reach the equivalence point. At the equivalence point, the moles of \(\text{NaOH}\) will equal the moles of \(\text{HClO}\).
\[
\text{moles of NaOH} = 0.0143834 \, \text{mol}
\]
Use the concentration of \(\text{NaOH}\) to find the volume required to reach the equivalence point:
\[
\text{volume of NaOH (L)} = \frac{\text{moles of NaOH}}{\text{concentration of NaOH}}
\]
Given:
- Concentration of \(\text{NaOH} = 0.306 \, \text{M}\)
\[
\text{volume of NaOH (L)} = \frac{0.0143834 \, \text{mol}}{0.306 \, \text{M}} = 0.0470 \, \text{L} = 47.0 \, \text{mL}
\]
At the equivalence point, the solution contains the salt sodium hypochlorite (\(\text{NaClO}\)), which hydrolyzes in water to form \(\text{ClO}^-\) and \(\text{OH}^-\). The reaction is:
\[
\text{ClO}^- + \text{H}_2\text{O} \rightleftharpoons \text{HClO} + \text{OH}^-
\]
The equilibrium constant for this reaction is the base dissociation constant \(K_b\), which can be calculated from the \(K_w\) and the \(K_a\) of \(\text{HClO}\):
\[
K_b = \frac{K_w}{K_a}
\]
Assume \(K_a\) for \(\text{HClO} = 3.0 \times 10^{-8}\) (a typical value) and \(K_w = 1.0 \times 10^{-14}\):
\[
K_b = \frac{1.0 \times 10^{-14}}{3.0 \times 10^{-8}} = 3.3333 \times 10^{-7}
\]
The concentration of \(\text{ClO}^-\) at the equivalence point is:
\[
[\text{ClO}^-] = \frac{\text{moles of ClO}^-}{\text{total volume (L)}}
\]
Total volume = \(29.8 \, \text{mL} + 47.0 \, \text{mL} = 76.8 \, \text{mL} = 0.0768 \, \text{L}\)
\[
[\text{ClO}^-] = \frac{0.0143834 \, \text{mol}}{0.0768 \, \text{L}} = 0.1873 \, \text{M}
\]
Using the \(K_b\) expression:
\[
K_b = \frac{[OH^-]^2}{[\text{ClO}^-]}
\]
\[
3.3333 \times 10^{-7} = \frac{x^2}{0.1873}
\]
Solving for \(x\) (concentration of \(\text{OH}^-\)):
\[
x^2 = 3.3333 \times 10^{-7} \times 0.1873
\]
\[
x^2 = 6.2433 \times 10^{-8}
\]
\[
x = \sqrt{6.2433 \times 10^{-8}} = 7.8996 \times 10^{-4} \, \text{M}
\]
Calculate the pOH:
\[
\text{pOH} = -\log(7.8996 \times 10^{-4}) = 3.103
\]
Finally, calculate the pH:
\[
\text{pH} = 14 - \text{pOH} = 14 - 3.103 = 10.897
\]
\[
\boxed{\text{pH} = 10.90}
\]
Answered
