Find out the osmotic pressure iof the 0.1m monobasic acid if ph is 2 at 25 degree celcius
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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
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