Question

Ph of a 0.1 m monobasic acid is found to be 2

Answer 1

Answer:

According to Arrhenius principle,

pH = - log[H⁺]

2 = -log[H⁺] ⇒[H⁺] = 10⁻² = 0.01

Assume HX is monobasic acid.

HX ⇄ H⁺ + X⁻

so, for dilute solution,

[H⁺ ] = Cα, here C is concentration of acid and α is dissociation Constant

Now, 0.01 = 0.1 × α [ ∵ concentration = 0.1M is given ]

α = 0.1

Now, A/C to van Hoff's principle,

i = 1 + α

i = 1 + 0.1 = 1.1

use Osmotic pressure, π = iCRT

= 1.1 × 0.1RT

= 0.11RT

Hence, answer is 0.11RT

Answer 2

Answer:

According to Arrhenius principle,

pH = - log[H⁺]

2 = -log[H⁺] ⇒[H⁺] = 10⁻² = 0.01

Assume HX is monobasic acid.

HX ⇄ H⁺ + X⁻

so, for dilute solution,

[H⁺ ] = Cα, here C is concentration of acid and α is dissociation Constant

Now, 0.01 = 0.1 × α [ ∵ concentration = 0.1M is given ]

α = 0.1

Now, A/C to van Hoff's principle,

i = 1 + α

i = 1 + 0.1 = 1.1

use Osmotic pressure, π = iCRT

= 1.1 × 0.1RT

= 0.11RT

Hence, answer is 0.11RT