Calculate the degree of hydrolysis of 0 . 005 M K 2 CrO 4 . K 2 = 5 . 0 × 10 - 7 for H 2 CrO 4 (It is essentially strong for first ionization).

K 2 CrO 4 ionizes as follows, K 2 CrO 4 → 2 K + + CrO 4 2 - CrO 4 2 - + H 2 O ( l ) ⇌ HCrO 4 - + OH - CrO 4 2 - + H 2 O ( l ) ⇌ HCrO 4 - + OH - C - Ch Ch Ch C is the concentration of K 2 CrO 4 and ' h ' is the degree of hydrolysis. For the above equilibrium, hydrolysis constant K h can be expressed as follows K h = Ch · Ch C - Ch ≈ Ch 2 HCrO 4 - is a weak acid and it's the conjugate base is CrO 4 2 - . Hence we can write K a × K h = K W at any given temperature K h = K W K a Given that C = 0 . 05 M and K a = K 2 = 5 × 10 - 7 At 25 o C we can write the hydrolysis constant as K h = 10 - 14 5 × 10 - 7 = 2 × 10 - 8 Now, h = K h C = 2 × 10 - 8 0 . 005 = 2 × 10 - 3

What is the pH of 0 . 1 M NaHCO 3 ? K 1 = 4 . 5 × 10 - 7 , K 2 = 4 . 5 × 10 - 11 for carbonic acids. (Given: log 2 = 0 . 3 , log 3 = 0 . 48 )

NaHCO 3 can acts as acid as well as a base. The following equilibria possible in the water HCO 3 - ⇌ H + + CO 3 2 - C - Cα Cα Cα The equilibrium constant of the above reaction = K a 2 K a 2 = H + CO 3 2 - HCO 3 - C is the concentration of the given salt and α is the degree of dissociation of it Bicarbonate can also undergo anionic hydrolysis HCO 3 - + H 2 O ⇌ H 2 CO 3 + OH - C - Ch Ch Ch C is the concentration of the salt and ' h ' is the degree of hydrolysis K h = H 2 CO 3 OH - HCO 3 - = K W K a 1 = H + OH - K a 1 Assume α ≃ h ( as both are small), then H 2 CO 3 = CO 3 2 - Now write K a 2 K h = H + CO 3 2 - HCO 3 - × HCO 3 - H 2 CO 3 OH - = K a 1 × K a 2 K W ⇒ H + OH - = K a 1 K a 2 K W = K a 1 K a 2 H + OH - ⇒ H + 2 = K a 1 × K a 2 ⇒ H + = K a 1 × K a 2 ⇒ pH = pK a 1 + pK a 2 2 pK a 1 = - log 4 . 5 × 10 - 7 = 6 . 346 pK a 2 = - log 4 . 5 × 10 - 11 = 10 . 346 pH = 8 . 346

E M B I B E

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.2>Q 15

HARD

JEE Main

IMPORTANT

Earn 100

$100 mL$ of $0 .5 M$ hydrazoic acid $({HN}_{3}, K_{a} = 3 .6 x 10^{- 4})$ and $400 mL$ of $0 .1 M$ $(HOCN, K_{a} = 8 \times 10^{- 4})$ are mixed. Which of the following is (are) true for the final solution?

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Important Questions on Ionic Equilibrium

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.2>Q 16

0 .1 M {CH}_{3} COOH

is diluted at

25^{o} C (K_{a} = 1.8 \times 10^{- 5})

then which of the following will be found correct?

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.2>Q 17

Which of the following statements is/are not correct?

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part 1 JEE Main & Advanced>Chapter 7 - Ionic Equilibrium>DPP 7.2>Q 18

Which of the following solutions when added to $1 L$ of a $0.01 M {CH}_{3} COOH$ solution will cause no change in the degree of dissociation of ${CH}_{3} COOH$ and $pH$ of the solution? $K_{a} = 1 .6 \times 10^{- 5}$ for ${CH}_{3} COOH$