[Solved] The ionization constant of benzoic acid is 6.46×10−5 and Ksp for silver benzoate is 2.5×10−13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water? - Chemistry Part-I

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Chemistry Part-I

Equilibrium

The ionization constant of benzoic acid is 6.46 { 10 }^{ -5 }

Chemistry Part-I

Class 11

NCERT

Chapter 1: Some Basic Concepts of Chemistry

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Chapter 2: Structure of Atom

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Chapter 3: Classification of Elements and Periodicity in Properties

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Chapter 5: States of Matter

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Chapter 6: Thermodynamics and Thermochemistry

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Chapter 7: Equilibrium

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Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

The ionization constant of benzoic acid is

6.46 \times 10^{- 5}

and

K_{s p}

for silver benzoate is

2.5 \times 10^{- 13}

. How many times is silver benzoate more soluble in a buffer of

p H 3.19

compared to its solubility in pure water?

Reaction between nitrogen and oxygen takes place as follows:

2 N_{2} (g) + O_{3} (g) ⇌ 2 N_{2} O (g)

If a mixture of

0.482 mol

N_{2}

and

0.933 mol

O_{2}

is placed in a reaction vessel of volume

10 L

and allowed to form

N_{2} O

at a temperature for which

K_{c} - 2.0 \times 1 0^{- 37}

, determine the composition of the equilibrium mixture.

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Question

Medium

Solving time: 5 mins

The ionization constant of benzoic acid is $6.46 \times 10^{- 5}$ and $K_{s p}$ for silver benzoate is $2.5 \times 10^{- 13}$ . How many times is silver benzoate more soluble in a buffer of $p H 3.19$ compared to its solubility in pure water?

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Updated on: Aug 29, 2023

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Chemistry Part-I (NCERT)

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The ionization constant of benzoic acid is

6.46 \times 10^{- 5}

and

K_{s p}

for silver benzoate is

2.5 \times 10^{- 13}

. How many times is silver benzoate more soluble in a buffer of

p H 3.19

compared to its solubility in pure water?

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

Let

s

M be the solubility of silver benzoate in water.

[A g^{+}] = [P h CO O^{-}] = s M

K_{s p} = [P h CO O^{-}] [A g^{+}] = s \times s = s^{2} = 2.5 \times 1 0^{- 13}

s = 5.0 \times 1 0^{- 7} M

. This is solubility of silver benzoate in pure water.

p H = - l o g [H^{+}] = 3.19

[H^{+}] = 6.456 \times 1 0^{- 4}

Benzoate ions combine with protons to form benzoic acid but hydrogen ion concentration is constant due to buffer solution.

P h COO H ⇌ P h CO O^{-} + H^{+}

K_{a} = \frac{[ P h CO O ^{-} ] [ H ^{+} ]}{[ P h COO H ]}

\frac{[ P h COO H ]}{[ P h CO O ^{-} ]} = \frac{[ H ^{+} ]}{K _{a}} = \frac{6.456 \times 1 0 ^{- 4}}{6.46 \times 1 0 ^{- 5}} = 10

Let y M be the soubiity in the buffer solution.

y = [A g^{+}] = [P h CO O^{-}] + [P h COO H] = [P h CO O^{-}] + 10 [P h CO O^{-}] = 11 [P h CO O^{-}]

[P h CO O^{-}] = \frac{y}{11}

K_{s p} = [P h CO O^{-}] [A g^{+}]

2.5 \times 1 0^{- 13} = \frac{y}{11} \times y

y = 1.66 \times 1 0^{- 6}

\frac{y}{s} = \frac{1.66 \times 1 0 ^{- 6}}{5.0 \times 1 0 ^{- 7}} = 3.32

Thus, silver benzoate is 3.32 times more soluble in buffer.

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Practice more questions from Chemistry Part-I (NCERT)

Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

The ionization constant of benzoic acid is

6.46 \times 10^{- 5}

and

K_{s p}

for silver benzoate is

2.5 \times 10^{- 13}

. How many times is silver benzoate more soluble in a buffer of

p H 3.19

compared to its solubility in pure water?

Reaction between nitrogen and oxygen takes place as follows:

2 N_{2} (g) + O_{3} (g) ⇌ 2 N_{2} O (g)

If a mixture of

0.482 mol

N_{2}

and

0.933 mol

O_{2}

is placed in a reaction vessel of volume

10 L

and allowed to form

N_{2} O

at a temperature for which

K_{c} - 2.0 \times 1 0^{- 37}

, determine the composition of the equilibrium mixture.

View all

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Question Text	The ionization constant of benzoic acid is $6.46 \times 10^{- 5}$ and $K_{s p}$ for silver benzoate is $2.5 \times 10^{- 13}$ . How many times is silver benzoate more soluble in a buffer of $p H 3.19$ compared to its solubility in pure water?
Updated On	Aug 29, 2023
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 7
Upvotes	694
Avg. Video Duration	11 min

The ionization constant of benzoic acid is $6.46 \times 10^{- 5}$ and $K_{s p}$ for silver benzoate is $2.5 \times 10^{- 13}$ . How many times is silver benzoate more soluble in a buffer of $p H 3.19$ compared to its solubility in pure water?

Text solutionVerified

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The ionization constant of benzoic acid is 6.46×10−5 and Ksp​ for silver benzoate is 2.5×10−13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Text solutionVerified

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The ionization constant of benzoic acid is $6.46 \times 10^{- 5}$ and $K_{s p}$ for silver benzoate is $2.5 \times 10^{- 13}$ . How many times is silver benzoate more soluble in a buffer of $p H 3.19$ compared to its solubility in pure water?