Question

0.1 M solution of weak monobasic acid (HX) at 27°C
has osmotic pressure 3.6 atm. The pH of 0.1 M
solution of HX at 27°C is??​

Answer 1

Answer:

1

Explanation:

PH=-log[H+]

[H+]=.1

PLEASE FOLLOW AND MARK AS BRAINLEST

Answer 2

Given:

C = 0.1 M

T = 27 °C = 300 K

Π = 3.6 atm

To Find:

The pH of the given monobasic acid.

Calculation:

- We know that:

Π = iCRT

⇒ 3.6 = i × 0.1 × 0.082 × 300

⇒ i = 3.6 / 2.46

⇒ i = 1.463

- For monobasic acid, n = 2

- Now, we know the relation:

α = (i - 1)/(n-1)

⇒ α = (1.463 - 1)/ (2 - 1)

⇒ α = 0.463

- As we know that:

[H⁺] = α²c/ (1 - α)

⇒ [H⁺] = 0.2144 × 0.1 /(1 - 0.463)

⇒ [H⁺] = 0.02144/0.537

⇒ [H⁺] = 0.04

- Now, the pH can be calculated as:

pH = - log [H⁺]

⇒ pH = - log (0.04)

⇒ pH = - log 4 × 10⁻²

⇒ pH = - (log 2² + log 10⁻²)

⇒ pH = - (2 log 2 - 2 log 10)

⇒ pH = - 2 (0.3010 - 1)

⇒ pH = - 2 × - 0.699

⇒ pH = 1.398 ≈ 1.4

- So, the pH of the given monobosic acid is about 1.4.