Questions: A 25.0 mL sample of 0.150 M hypochlorous acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of hypochlorous acid is 3.0 x 10^-8. 14.52 10.20 7.00 6.48 7.52

Solution Steps

At the equivalence point, all the hypochlorous acid (HClO) has reacted with NaOH to form its conjugate base, sodium hypochlorite (NaClO). The reaction is:

\[ \text{HClO} + \text{NaOH} \rightarrow \text{NaClO} + \text{H}_2\text{O} \]

Since the volumes and concentrations of HClO and NaOH are equal, the concentration of the conjugate base (ClO\(^-\)) at the equivalence point is the same as the initial concentration of HClO, which is 0.150 M.

The \(K_b\) of the conjugate base can be found using the relation:

\[ K_w = K_a \times K_b \]

where \(K_w = 1.0 \times 10^{-14}\). Given \(K_a = 3.0 \times 10^{-8}\), we find:

\[ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{3.0 \times 10^{-8}} = 3.3333 \times 10^{-7} \]

The concentration of OH\(^-\) ions can be calculated using the expression for \(K_b\):

\[ K_b = \frac{[OH^-]^2}{[ClO^-]} \]

Substituting the known values:

\[ 3.3333 \times 10^{-7} = \frac{[OH^-]^2}{0.150} \]

Solving for \([OH^-]\):

\[ [OH^-]^2 = 3.3333 \times 10^{-7} \times 0.150 = 5.0000 \times 10^{-8} \]

\[ [OH^-] = \sqrt{5.0000 \times 10^{-8}} = 7.0711 \times 10^{-4} \]

First, calculate the pOH:

\[ \text{pOH} = -\log(7.0711 \times 10^{-4}) = 3.1505 \]

Then, calculate the pH:

\[ \text{pH} = 14 - \text{pOH} = 14 - 3.1505 = 10.8495 \]

Final Answer