Question

Given that Ka for HBrO is 2.8 × 10-9 at 25 °C, what is the value...

Given that Ka for HBrO is 2.8 × 10-9 at 25 °C, what is the value of Kb for BrO– at 25 °C?

Given that Kb for (CH3)3N is 6.3 × 10-5 at 25 °C, what is the value of Ka for (CH3)3NH at 25 °C?

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Homework Answers

Answer #1

Solution:

So calculate value of BrO-, We use ionic product of water.

Kw = 1.0 E-14

Value of ka and kb for its conjugate base is related by following expression.

Ka x kb = 1.0 E-14

We know the ka and by using it we can get the value of kb of its conjugate base.

Kb = 1.0 E-14 / ka

            =1.0 E-14 / 2.8E-9

            =3.57 E-6

b)

calculation of ka of (CH3)3NH

We use same expression to calculate the value of kb

Ka = 1.0 E-14 / 6.3 E-5

= 1.59 E-10

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