Question

Osmotic pressure at T temperature of CH3COOH 0.5 M aqueous solution containing 2 pH​

Answer 1

Explanation: CH3COOH is a mono basic acid.

CH3COOH⇌CH3COO- + H+

Relation b/w[H+] & degree of dissociation

[H+]=cα

So α=[H+]/c

Therefore α=0.01/0.5=0.02

Now, According to vant Hoff's principle,

i=1+α

i=1+0.02⇒i=1.02

Osmotic pressure ∏=iCRT

∏=1.02×0.5×RT

∏=0.51RT

Answer 2

Answer:

The osmotic pressure, π, of the solution measured is $0.51RT.$

Explanation:

Given,

The concentration of the solution, C = $0.5M$

The temperature = T

The pH = $2$

The osmotic pressure, π =?

As we know,

• Osmotic pressure,  π = $i CRT$      -------equation (1)

Here,

• i = van't Hoff factor
• C = concentration
• R = Raydberg's constant
• T = temperature

As given, pH = 2

Therefore, $[H^{+}]$ = $10^{-pH}$ = $10^{-2}$ = $0.01M$

Firstly, we have to calculate the value of the van't Hoff factor.

• i = $1 + \alpha$    -------equation (2)

Here, $\alpha$ is the degree of dissociation.

Now, the dissociation of  $CH_{3}COOH$ occurs in the way given below:

• $CH_{3}COOH \rightarrow CH_{3}COO^{-} + H^{+}$

From this reaction, we can say, $CH_{3}COOH$ is a monobasic acid.

Also,

• $[H^{+}]$ = $C\alpha$
• $0.01$ = $0.5\alpha$
• $\alpha = 0.02$

After putting the value of $\alpha$ in equation (2), we get:

• i = $1+ 0.02$ = $1.02$

After putting the values in equation (1), we get:

• Osmotic pressure, π = $1.02 \times 0.5 RT$ = $0.51RT$.