At T ( K ) , if the ionization constants of HCOOH and HCN are 2 . 0 × 10 - 4 and 5 . 0 × 10 - 9 , respectively, the ionization constants of the corresponding conjugate bases, respectively, are

5 . 0 × 10 - 11 , 2 . 0 × 10 - 6

2 . 0 × 10 - 6 , 5 . 0 × 10 - 11

2 . 0 × 10 - 4 , 5 . 0 × 10 - 9

5 . 0 × 10 - 9 , 2 . 0 × 10 - 4

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

EASY

Earn 100

$pOH$ of water is $7.0 at 298 K$ . If water is heated to $350 K$ , which of the following would be true?

43.59% studentsanswered this correctly

Important Questions on Equilibrium

EASY

Chemistry>Physical Chemistry>Equilibrium>Ionic Product of Water

The

pH

of pure water at

80 ° C

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>Ionic Product of Water

The

pH

of an aqueous solution of

10^{- 8} MHCl

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>Ionic Product of Water

T (K)

, if the ionization constants of

HCOOH

and

HCN

are

2.0 \times 10^{- 4}

and

5.0 \times 10^{- 9}

, respectively, the ionization constants of the corresponding conjugate bases, respectively, are

HARD

Chemistry>Physical Chemistry>Equilibrium>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.

HARD

Chemistry>Physical Chemistry>Equilibrium>Ionic Product of Water

298 K,

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

EASY

Chemistry>Physical Chemistry>Equilibrium>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) ${NH}_{3 (aq)} + {OH}_{(aq)}^{-} ⇆ {NH}_{2 (aq)}^{-} + H_{2} O$

EASY

Chemistry>Physical Chemistry>Equilibrium>Ionic Product of Water

The

pH

of pure water at

90 {}^{\circ}C

will be

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>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

EASY

Chemistry>Physical Chemistry>Equilibrium>Ionic Product of Water

Explain the term ‘ionic product of water’.

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>Ionic Product of Water

At what molar concentration of HCl will its aqueous solution have an [H⁺] to which equal contribution come from HCl and H₂O at 90^oC.

[K_w of H₂O = 10^-12 M² at 90^oC]

MEDIUM

Chemistry>Physical Chemistry>Equilibrium>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: