Question

A $0.01M$ aqueous solution of weak acid $HA$ has an osmotic pressure $0.293$ atm at ${25}^{o}C$. Another $0.01M$ aqueous solution of other weak acid $HB$ has an osmotic pressure of $0.345$ atm under the same conditions. Equilibrium constants of second acid for their dissociation is $X\times {10}^{-3}$, nearest integer to X is:

Answer 1

$\underset{11-\alpha }{HA}\rightleftharpoons \underset{0\alpha }{H^{+}}+\underset{0\alpha }{A^{-}}$
Now,

$\pi V=nST(1+\alpha )$

or $1+\alpha =\frac{\pi \times V}{n\times S\times T}$

or $1+\alpha =\frac{0.293}{0.01\times 0.0821\times 298}(\frac{n}{V}=0.01)$

$\therefore \alpha =1.197-1=0.197$

$\therefore K_a=\frac{c{\alpha }^2}{(1-\alpha )}=\frac{0.01\times {\left(0.197\right)}^2}{(1-0.197)}=4.83\times {10}^{-4}$

Similarly, $K_a$ for $HB=2.85\times {10}^{-3}$