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Chemistry Part-I

Equilibrium

The pH of 0.005 M codeine ( { C }_{ 18 }{ H }_{ 21 }{ NO }_{

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

p H

of the following solutions:

a) 2 g of

TlO H

dissolved in water to give 2 litre of solution.
b) 0.3 g of

C a (O H)_{2}

dissolved in water to give 500 mL of solution.
c) 0.3 g of

N a O H

dissolved in water to give 200 mL of solution.
d) 1 mL of 13.6 M

H Cl

is diluted with water to give 1 litre of solution.

The

p H

of 0.005 M codeine (

C_{18} H_{21} NO_{3}

) solution is 9.95. Calculate its ionization constant and

p K_{b}

What is the

p H

of 0.001 M aniline solution ? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.

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Question

Medium

Solving time: 5 mins

The $p H$ of 0.005 M codeine ( $C_{18} H_{21} NO_{3}$ ) solution is 9.95. Calculate its ionization constant and $p K_{b}$ .

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Chemistry Part-I (NCERT)

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The

p H

of 0.005 M codeine (

C_{18} H_{21} NO_{3}

) solution is 9.95. Calculate its ionization constant and

p K_{b}

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Text solutionVerified

p H = 9.95

pO H = 14 - p H = 14 - 9.95 = 4.05

[O H^{-}] = 1 0^{- pO H} = 1 0^{- 4.05} = 8.913 \times 1 0^{- 5}

C o d e in e + H_{2} O ⇌ C o d e in e H^{+} + O H^{-}

The ionization constant,

K_{b} = \frac{[ C o d e in e H ^{+} ] [ O H ^{-} ]}{[ co d e in e ]} = \frac{( 8.913 \times 1 0 ^{- 5} ) \times ( 8.913 \times 1 0 ^{- 5} )}{5 \times 1 0 ^{- 3}} = 1.588 \times 1 0^{- 6}

p K_{b} = - l o g (1.588 \times 1 0^{- 6}) = 5.8

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Practice more questions from Chemistry Part-I (NCERT)

Calculate the

p H

of the following solutions:

a) 2 g of

TlO H

dissolved in water to give 2 litre of solution.
b) 0.3 g of

C a (O H)_{2}

dissolved in water to give 500 mL of solution.
c) 0.3 g of

N a O H

dissolved in water to give 200 mL of solution.
d) 1 mL of 13.6 M

H Cl

is diluted with water to give 1 litre of solution.

The

p H

of 0.005 M codeine (

C_{18} H_{21} NO_{3}

) solution is 9.95. Calculate its ionization constant and

p K_{b}

What is the

p H

View all

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Question Text	The $p H$ of 0.005 M codeine ( $C_{18} H_{21} NO_{3}$ ) solution is 9.95. Calculate its ionization constant and $p K_{b}$ .
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1
Upvotes	74

The $p H$ of 0.005 M codeine ( $C_{18} H_{21} NO_{3}$ ) solution is 9.95. Calculate its ionization constant and $p K_{b}$ .

Text solutionVerified

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The $p H$ of 0.005 M codeine ( $C_{18} H_{21} NO_{3}$ ) solution is 9.95. Calculate its ionization constant and $p K_{b}$ .