Given that Ka for HClO is 4.0 × 10-8 at 25°C, what is the value of Kb for ClO– at 25°C?

The question is –

Given that K_a for HClO is 4.0 × 10^-8 at 25°C, what is the value of K_b for ClO^– at 25°C?

Answer:

⇒The value of K_b for ClO^– at 25 °C is 2.5 x 10^-7.

Explanation:

Step 1: The first step is to use the relationship between K_a and K_b for a conjugate acid-base pair, which is given by:

K_b x K_a = K_w   ………….equation (1)

⇒ K_a is the acid dissociation constant, which is a measure of the strength of an acid in a solution.

⇒ K_b is the base dissociation constant, which is a measure of the strength of a base in a solution.

⇒ K_w is the ion product constant of water at 25 °C, which is 1.0 x 10^-14.

Step 2: Next, we need to substitute the given K_a value for HClO (4.0 x 10^-8) and K_w in equation 1 from step 1:

∴ K_b x K_a = K_w

⇒ K_b x 4.0 x 10^-8 = 1.0 x 10^-14

Step 3: Solve for K_b by dividing both sides of the equation by the K_a value:

⇒ K_b = 1.0 x 10^-14/4.0 x 10^-8

Step 4: Now we can calculate the value of K_b for ClO^– at 25 °C:

∴ K_b = 2.5 x 10^-7

So, the final answer is that the value of K_b for ClO^– at 25°C is 2.5 x 10^-7.

Also read: 

⇒ How to find K_b from K_a? – With examples.