Question

The osmotic pressure of 0.5M aqueous solution of acetic acid having pH=2 at temperature T is​

Answer 1

Explanation:

pH=2

According to Arrhenius principle.

pH=−log[H

+

]

2=−log[H

+

]

[H

+

]=0.01.

Assume HX is a mono basic acid.

HX⇌H

+

+X

−

Relation b/w[H

+

] & degree of dissociation

[H

+

]=cα

0.1

0.01

=α

α⇒0.1

Now, According to vant Hoff's principle,

i=1+α

i=1+0.1⇒i=1.1

Osmotic pressure ∏=iCRT

∏=1.1×0.1×RT

∏=0.11RT

Answer 2

Given:

0.5M aqueous solution of acetic acid having pH=2 at temperature T.

To find:

Osmotic pressure.

Calculation:

$\therefore$ pH = 2

$\{ {H}^{ + } \} = {10}^{ - 2} \: M$

Now , Let degree of dissociation be $\alpha$.

$\alpha = \dfrac{ {10}^{ - 2} }{0.5} = 2 \times {10}^{ - 2}$

Now , Van't Hoff Factor be i;

$i = \alpha + 1$

$= > i = (2 \times {10}^{ - 2} ) + 1$

$= > i = (0.02) + 1$

$= > i = 1.02$

Let osmotic pressure be $\pi$

$\therefore \: \pi =i \times c \times R \times T$

$= > \: \pi =1.02\times 0.5 \times R \times T$

$= > \: \pi =0.51 R T$

So, final answer is:

$\boxed{ \sf{\: \pi =0.51 R T}}$