Upon passing 0 . 01 mole HCl gas through 100 mL of 0 . 05 M Formic acid solution K a = 1 . 8 × 10 - 4 , determine change in pH of solution and [ HCOO - ] in resulting solution.

Formic acid formula is HCOOH and it is a weak acid. Its pK a value can be calculated as follow- pK a = - logK a = - log 1 . 8 × 10 - 4 = 3 . 74 Its pH formula is only when HCOOH dissociates- HCOOH ⇌ H + + HCOO - pH = 1 2 pK a - logC = 1 2 3 . 74 - log 0 . 05 = 1 2 3 . 74 - - 1 . 3 = 2 . 52 ......equation 1 On passing 0 . 01 mole HCl gas through 100 ml of 0 . 05 M formic acid- Molarity of HCl = moles of HCl volume of solution in litre = 0 . 01 0 . 1 = 0 . 1 M HCl → H + + Cl - t = 0 0 . 1 M t = t 0 M 0 . 1 M 0 . 1 M HCOOH ⇌ H + + HCOO - , K a = H + HCOO - HCOOH t = 0 0 . 05 M 0 . 1 M 0 M t = t 0 . 05 - x M 0 . 1 + x M x M K a = 1 . 8 × 10 - 4 = 0 . 1 + x x 0 . 05 - x As K a value of HCOOH is very small and common ion effect due to HCl is acting on it which reduce its dissociation more. So value of x is very small. We can neglect value of x with respect to 0 . 05 M and 0 . 1 M . K a = 1 . 8 × 10 - 4 ≃ 0 . 1 x 0 . 05 = 2 x x = 1 . 8 × 10 - 4 2 = 9 × 10 - 5 M H + = 0 . 1 + x M ≃ 0 . 1 M New pH = - log H + = - log 0 . 1 = 1 ......equation 2 Change in pH = 2 . 52 - 1 = 1 . 52 [ HCOO - ] = x M = 9 × 10 - 5 M

Calculate H + , HCOO - and OCN - in a solution that contains 0 . 1 M HCOOH K a = 2 . 4 × 10 - 4 and 0 . 1 M HOCN K a = 4 × 10 - 4 .

When two weak acids are mixed to obtain an aqueous solution, it becomes a simple case of simultaneous equilibria in which both the weak acids undergo dissociation simultaneously. Here, HCOOH and HOCN , both are weak acids. Let us assume that the concentrations and degree of dissociation of HCOOH and HOCN be C 1 and α 1 and α 1 , respectively. HCOOH ⇌ H + + COO - t = 0 C 1 0 0 t = t eq C 1 - C 1 α 1 C 1 α 1 + C 2 α 2 C 1 α 1 K a 1 = ( C 1 α 1 ) ( C 1 α 1 + C 2 α 2 ) ( C 1 - C 1 α 1 ) K a 1 ( C 1 - C 1 α 1 ) = ( C 1 α 1 + C 2 α 2 ) ( C 1 α 1 ) - - - - - - ( 1 ) HOCN ⇌ H + + OCN - t = 0 C 2 0 0 t = t eq C 2 - C 2 α 1 C 1 α 1 + C 2 α 2 C 2 α 2 K a 2 = ( C 2 α 2 ) ( C 1 α 1 + C 2 α 2 ) ( C 2 - C 2 α 2 ) K a 2 ( C 2 - C 2 α 2 ) = ( C 1 α 1 + C 2 α 2 ) ( C 2 α 2 ) - - - - - - ( 2 ) Adding 1 and 2, we get: K a 1 ( C 1 - C 1 α 1 ) + K a 2 ( C 2 - C 2 α 2 ) = ( C 1 α 1 + C 2 α 2 ) 2 [ H + ] = ( C 1 α 1 + C 2 α 2 ) = K a 1 ( C 1 - C 1 α 1 ) + K a 2 ( C 2 - C 2 α 2 ) = K a 1 C 1 ( 1 - α 1 ) + K a 2 C 2 ( 1 - α 2 ) Ignoring α 1 and α 2 , with respect to 1 , we get: [ H + ] = ( C 1 α 1 + C 2 α 2 ) = K a 1 C 1 + K a 2 C 2 In the question, K a 1 = 2 . 4 × 10 - 4 , C 1 = 0 . 1 M , K a 2 = 4 × 10 - 4 , C 2 = 0 . 1 M [ H + ] = ( 0 . 1 × 2 . 4 × 10 - 4 + 0 . 1 × 4 × 10 - 4 ) = 0 . 8 × 10 - 2 M K a 1 = ( 0 . 8 × 10 - 2 ) ( 0 . 1 α 1 ) 0 . 1 ( 1 - α 1 ) ⇒ 2 . 4 × 10 - 4 = ( 0 . 8 × 10 - 2 ) ( 0 . 1 α 1 ) 0 . 1 ( 1 - α 1 ) Ignoring α 1 with respect to 1 , we get: α 1 = 3 × 10 - 2 [ COO - ] = 0 . 1 α 1 = 0 . 3 × 10 - 2 K a 2 = ( 0 . 8 × 10 - 2 ) ( 0 . 1 α 2 ) 0 . 1 ( 1 - α 2 ) ⇒ 4 × 10 - 4 = ( 0 . 8 × 10 - 2 ) ( 0 . 1 α 2 ) 0 . 1 ( 1 - α 2 ) Ignoring α 1 with respect to 1 , we get: α 2 = 5 × 10 - 2 [ OCN - ] = 0 . 1 α 2 = 0 . 5 × 10 - 2

The exothermic formation of ClF 3 is represented by the following equation: Cl 2 g + 3 F 2 g ⇌ 2 ClF 3 g ; ΔH = - 329 kJ Which of the following will increase the quantity of ClF 3 in the equilibrium mixture of Cl 2 , F 2 and ClF 3 ?

Increasing the volume of the container.

E M B I B E

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 197

EASY

JEE Main/Advance

IMPORTANT

Earn 100

Upon mixing equal volume of a strong acid solution ( $HA$ ) and a strong base $(BOH)$ solution, $pH$ of resulting solution:

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Important Questions on Equilibrium

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 198

Which of the following solutions will have

pH

close to

1

?

EASY

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 199

Which of the following has the maximum degree of ionisation?

EASY

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 200

K_{a}

for formic acid and acetic acid are

1.8 \times 10^{- 4}

and

1.8 \times 10^{- 5}

respectively. The relative strength of acids is:

MEDIUM

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 201

K_{a}

for a monobasic acid, whose

0.1 M

solution has

pH

of

4.5

is:

EASY

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 204

The

p H

of 0.1 M solution of the following salts increases in an order -

MEDIUM

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 205

The degree of hydrolysis of a salt of weak monobasic acid and weak monoacidic base in its $0.1 M$ solution is found to be $50 %$ . If the molarity of the solution is $0.2 M,$ the percentage hydrolysis of the salt should be: