Question

What is the osmotic pressure of 0.1 M $HA$ (aq.) solution at $300K$? (Given, $K_{\left(HA\right)}$ = ${10}^{-6}$)

• A. $1.23atm$
• B. $0.52atm$
• C. $2.48atm$
• D. $3.5atm$

Answer 1

Answer: (C)

$2.48atm$

Osmotic pressure $\left(\pi \right)=\text{i}\text{CRT}$
$\underset{0.1\left(1-\alpha \right)}{\underset{0.1}{\text{HA}}}\rightleftharpoons \underset{0.1\alpha }{\underset{0}{{\text{H}}^{+}}}+\underset{0.1\alpha }{\underset{0}{{\text{A}}^{-}}}$
$\text{K}={10}^{-6}=\frac{{\left(0.1\alpha \right)}^2}{0.1\left(1-\alpha \right)}=\frac{0.1{\alpha }^2}{1-\alpha }$
Here,
$1>>\alpha$
${10}^{-6}=0.1{\alpha }^2$
${\alpha }^2={10}^{-5}$ (since, $\text{i}=1+\alpha$, i = 1.0032)
Now, $\pi ={\left({10}^{-5}\right)}^{1/2}\times 0.1\times 0.0821\times 300$
$\pi =1.003\times 0.1\times 0.0821\times 300$
$\pi =2.48\text{atm}$
Hence, Option (C) is correct.