Questions: Complete the table below. Round each of your entries to 2 significant digits. You may conjugate acid conjugate base ------------ formula Ka formula Kb 5.9 × 10^-6 C5H5N 1.7 × 10^-9 3.1 × 10^-8 ClO^- 3.3 × 10^-9 4.59 × 10^-7 HCO3 2.2 × 10^-8

Solution Steps

The relationship between the acid dissociation constant (\(K_a\)) and the base dissociation constant (\(K_b\)) for a conjugate acid-base pair is given by the equation:

\[ K_a \times K_b = K_w \]

where \(K_w\) is the ion-product constant of water, approximately \(1.0 \times 10^{-14}\) at 25°C.

For each conjugate base given, we need to find the corresponding conjugate acid. The conjugate acid of a base is formed by adding a proton (\(H^+\)) to the base.

1. Conjugate Base: \(\mathrm{C}_5\mathrm{H}_5\mathrm{N}\)

   • Conjugate Acid: \(\mathrm{C}_5\mathrm{H}_5\mathrm{NH}^+\)

2. Conjugate Base: \(\mathrm{ClO}^-\)

   • Conjugate Acid: \(\mathrm{HClO}\)

3. Conjugate Base: \(\mathrm{HCO}_3^-\)

   • Conjugate Acid: \(\mathrm{H}_2\mathrm{CO}_3\)

Using the relationship \(K_a \times K_b = K_w\), we can verify the given \(K_a\) and \(K_b\) values:

1. For \(\mathrm{C}_5\mathrm{H}_5\mathrm{NH}^+\) and \(\mathrm{C}_5\mathrm{H}_5\mathrm{N}\):

   \[ K_a = 5.9 \times 10^{-6}, \quad K_b = 1.7 \times 10^{-9} \]

   \[ K_a \times K_b = (5.9 \times 10^{-6}) \times (1.7 \times 10^{-9}) = 1.003 \times 10^{-14} \approx K_w \]

2. For \(\mathrm{HClO}\) and \(\mathrm{ClO}^-\):

   \[ K_a = 3.1 \times 10^{-8}, \quad K_b = 3.3 \times 10^{-9} \]

   \[ K_a \times K_b = (3.1 \times 10^{-8}) \times (3.3 \times 10^{-9}) = 1.023 \times 10^{-16} \neq K_w \]

   (Note: There seems to be a discrepancy here, but we will proceed with the given values.)

3. For \(\mathrm{H}_2\mathrm{CO}_3\) and \(\mathrm{HCO}_3^-\):

   \[ K_a = 4.59 \times 10^{-7}, \quad K_b = 2.2 \times 10^{-8} \]

   \[ K_a \times K_b = (4.59 \times 10^{-7}) \times (2.2 \times 10^{-8}) = 1.0098 \times 10^{-14} \approx K_w \]

Final Answer