The ionization constant of H F , H C O O H and H C N at 298 K are 6.8 × 10 - 4 , 1.8 × 10 - 4 and 4.8 × 10 - 9 respectively. Calculate the ionization constants of the corresponding conjugate base.

The ionization constant of H F at 298 K is 6.8 × 10 - 4 The ionization constant of H C O O H at 298 K is 1.8 × 10 - 4 The ionization constant of H C N at 298 K is 4.8 × 10 - 9 Formula : K b K a = K w K b = K w K a K a of H F = 6.8 × 10 - 4 Hence, K b of its conjugate base F - = K w K a = 10 - 14 6.8 × 10 - 4 = 1. 47 × 10 - 11 K a of H C O O H = 1.8 × 10 - 4 Hence, K b of its conjugate base H C O O - = K w K a = 10 - 14 1.8 × 10 - 4 = 5 . 56 × 10 - 11 K a of H C N = 4.8 × 10 - 9 Hence, K b of its conjugate base C N - = K w K a = 10 - 14 4.8 × 10 - 9 = 2 . 08 × 10 - 6

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The ionic product of water if a few drops of acid or base are added to it.(remains constant/decreases/increases)

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Important Questions on Chemical Equilibrium and Acids-Bases

HARD

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

298 K,

the ratio of dissociated water to that of undissociated water is [Assume pure water]

MEDIUM

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

The

pH

of an aqueous solution of

10^{- 8} MHCl

HARD

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

For the following Assertion and Reason, the correct option is:
Assertion: The

p H

of water increases with increase in temperature.
Reason: The dissociation of water into

H^{+}

and

O H^{-}

is an exothermic reaction.

EASY

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

The

pH

of pure water at

80 ° C

EASY

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

The

pH

of pure water at

90 {}^{\circ}C

will be

EASY

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

Which of the following equations give ionic product of water?

(i) ${NH}_{3 (aq)} + H_{2} O ⇆ {NH}_{4 (aq)}^{+} + {OH}_{(aq)}^{-}$

(ii) ${NH}_{4 (aq)}^{+} + H_{2} O ⇆ H_{3} O^{+} + {NH}_{3 (aq)}$

(iii) ${NH}_{2 (aq)}^{-} + H_{2} O ⇆ {NH}_{3 (aq)} + {OH}^{-}$

(iv) ${NH}_{3 (aq)} + {OH}_{(aq)}^{-} ⇆ {NH}_{2 (aq)}^{-} + H_{2} O$

MEDIUM

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

How many litres of water must be added to

1

litre of aqueous solution of

HCl

with a pH of

1

to create an aqueous solution with pH of

2

MEDIUM

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

How does

K_{w}

vary with temperature and why?

EASY

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

By applying the law of mass action, the equilibrium constant,

K

for the reaction

HA + H_{2} O ⇌ H_{3} O^{+} + A^{-}

, is given as

MEDIUM

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

What do you understand by the term ‘ionic product of water’? How has this concept been useful in defining the acidity and basicity of a solution?

MEDIUM

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

0.1 mole of ${CH}_{3} \cdot {NH}_{2} (K_{b} = 5 \times 10^{- 4})$ is added to 0.08 moles of $HCl$ and solution is diluted to one liter. The resulting $[H^{+}]$ is:

MEDIUM

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

Explain the terms ‘Ionic product of water’ and ‘

pH

value’. How does the former change with temperature? What is

pH

scale?

EASY

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

The

pH

of a solution is

6

. Sufficient amount of acid is added to decrease the

pH

2

. The increase in hydrogen ion concentration is

EASY

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

Explain the term ‘ionic product of water’.

EASY

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

For pure water at

25^{o} C

and

50^{o} C

, the correct statement is

HARD

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

The ionization constant of

H F, H C O O H

and

H C N

298 K

are

6.8 \times 10^{- 4}, 1.8 \times 10^{- 4}

and

4.8 \times 10^{- 9}

respectively. Calculate the ionization constants of the corresponding conjugate base.

EASY

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

K_{a} = 10^{- 5}

for a weak acid, then

{p K}_{b}

for its conjugate base would be -

EASY

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

What is the [OH^–] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)₂ ?

MEDIUM

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

What do you understand by the term 'ionic product of water'? How is it represented?

EASY

Chemistry>Physical Chemistry>Chemical Equilibrium and Acids-Bases>Ionic Product of Water

80 {}^{0}C [H_{3} O^{+}]

, the concentration is equal to

1 \times 10^{- 6} mol L^{- 1}

. At the same temperature, the value of

K_{w}

is:

The ionic product of water if a few drops of acid or base are added to it.(remains constant/decreases/increases)

Important Questions on Chemical Equilibrium and Acids-Bases

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