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Chemistry Part-I

Equilibrium

The {pH} of 0.004 {M} hydrazine solution is 9.7. Calculate its

Chemistry Part-I

Class 11

NCERT

Chapter 1: Some Basic Concepts of Chemistry

44 questions

Chapter 2: Structure of Atom

85 questions

Chapter 3: Classification of Elements and Periodicity in Properties

50 questions

Chapter 4: Chemical Bonding and Molecular Structure

44 questions

Chapter 5: States of Matter

28 questions

Chapter 6: Thermodynamics and Thermochemistry

36 questions

Chapter 7: Equilibrium

101 questions

Calculate the

pH

0.08 M

solution of hypochlorous acid,

HOCl

. The ionization constant of the acid is

2.5 \times 1 0^{- 5}

. Determine the percent dissociation of

HOCl

The

pH

0.004 M

hydrazine solution is 9.7. Calculate its ionization constant

K_{b}

and

p K_{b}

Calculate the

pH

of the solution in which

0.2 M NH_{4} Cl

and

0.1 M NH_{3}

are present. The

pK_{b}

of ammonia solution is

4.75

View all question

Question

Medium

Solving time: 3 mins

The $pH$ of $0.004 M$ hydrazine solution is 9.7. Calculate its ionization constant $K_{b}$ and $p K_{b}$ .

Views: 5,651 students

Updated on: Oct 29, 2023

Book:

Chemistry Part-I (NCERT)

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The

pH

0.004 M

hydrazine solution is 9.7. Calculate its ionization constant

K_{b}

and

p K_{b}

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

Solution

NH_{2} NH_{2} + H_{2} O ⇌ NH_{2} NH_{3}^{+} + OH^{-}

From the

pH

we can calculate the hydrogen ion concentration. Knowing hydrogen ion concentration and the ionic product of water we can calculate the concentration of hydroxyl ions. Thus we have:

[H^{+}] = antilog (- pH)

= antilog (- 9.7) = 1.67 \times 1 0^{- 10}

[OH^{-}] = K_{w} / [H^{+}] = 1 \times 1 0^{- 14} /1.67 \times 1 0^{- 10}

= 5.98 \times 1 0^{- 5}

The concentration of the corresponding hydrazinium ion is also the same as that of hydroxyl ion. The concentration of both these ions is very small so the concentration of the undissociated base can be taken equal to

0.004 M

.
Thus,

K_{b} = [NH_{2} NH_{3}^{+}] [OH] / [NH_{2} NH_{2}]

= (5.98 \times 1 0^{- 5})^{2} /0.004 = 8.96 \times 1 0^{- 7}

p K_{b} = - lo g K_{b} = - lo g (8.96 \times 1 0^{- 7}) = 6.04

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Practice more questions from Chemistry Part-I (NCERT)

Calculate the

pH

0.08 M

solution of hypochlorous acid,

HOCl

. The ionization constant of the acid is

2.5 \times 1 0^{- 5}

. Determine the percent dissociation of

HOCl

The

pH

0.004 M

hydrazine solution is 9.7. Calculate its ionization constant

K_{b}

and

p K_{b}

Calculate the

pH

of the solution in which

0.2 M NH_{4} Cl

and

0.1 M NH_{3}

are present. The

pK_{b}

of ammonia solution is

4.75

View all

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Question Text	The $pH$ of $0.004 M$ hydrazine solution is 9.7. Calculate its ionization constant $K_{b}$ and $p K_{b}$ .
Updated On	Oct 29, 2023
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 3
Upvotes	339
Avg. Video Duration	15 min

The $pH$ of $0.004 M$ hydrazine solution is 9.7. Calculate its ionization constant $K_{b}$ and $p K_{b}$ .

Text solutionVerified

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The $pH$ of $0.004 M$ hydrazine solution is 9.7. Calculate its ionization constant $K_{b}$ and $p K_{b}$ .