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Chapter 6: Q5TY (page 138)

$K_{a}$ for chloroacetic acid is $1.36 脳 10^{- 3} .$ Find $K_{b}$ for chloroacetate ion.

Short Answer

Expert verified

In the chloroacetate ion the value of $K_{b} is 7.4 脳 10^{- 12}$

Step by step solution

01

Step 1: Define the formula for the calculation of Kb .

The value of $K_{b}$ can be calculated as:

$K_{b} = \frac{[{BH}^{+}] [OH]}{[B]}$

Where $K_{b}$ is called the base dissociation constant.

The value of $K_{w}$ is calculated by the formula:

$K_{w} = K_{a} 脳 K_{b}$

02

Calculate the value of Kb.

Given:

$K_{a} = 1.36 脳 10^{- 3}$

$K_{b}$ for chloroacetate ion

$K_{w} = K_{a} 脳 K_{b}$

Where,

$(K_{b})$ is the base dissociation constant.

$(K_{w})$ is the ionic product of water $(1.00 脳 10^{- 14})$

$(K_{a})$ is the acid dissociation constant

$K_{b} = \frac{K_{w}}{K_{a}} = \frac{1.0 脳 10^{- 14}}{1.36 脳 10^{- 3}} = = 0.735 脳 10^{- 11} = 7.4 脳 10^{- 12}$

Thus, in the chloroacetate ion, the value of role="math" localid="1663408479676" $K_{b} is 7.4 脳 10^{- 12}$

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