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

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## 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 value of Kb for ClO at 25 °C is 2.5 x 10-7.

Explanation:

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

Kb x Ka = Kw   ………….equation (1)

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

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

⇒ Kw 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 Ka value for HClO (4.0 x 10-8) and Kw in equation 1 from step 1:

∴ Kb x Ka = Kw

⇒ Kb x 4.0 x 10-8 = 1.0 x 10-14

Step 3: Solve for Kb by dividing both sides of the equation by the Ka value:

⇒ Kb = 1.0 x 10-14/4.0 x 10-8

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

∴ Kb = 2.5 x 10-7

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

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