Question

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

Answer 1

Hint:Equilibrium dissociation constant is used to measure the strength of that species in the solution. Solvent molecules always flow from region of lower concentration to region of higher concentration of solution. Osmotic pressure depends on the concentration of the solution.
Formula used: $i = \dfrac{{\pi V}}{{nRT}}$ and $i = 1 + \alpha $
Where i is Van’t Hoff factor, $\pi $ is osmotic pressure, $V$ is volume of the sample, $n$ is number of moles, $R$ is gas constant, $T$ is temperature and $\alpha $ is degree of dissociation.

Complete step by step answer:
Osmotic pressure is that pressure which stops the flow of solvent from a region of higher concentration to a region of lower concentration in a solution through a semipermeable membrane. Osmotic pressure is denoted by $\pi $ .
An acid dissociation constant is a measure of the strength of an acid in a solution. It is usually denoted by ${K_a}$ .
For weak acid HA, ${\pi _1} = 0.293atm$
For weak acid HB, ${\pi _2} = 0.345atm$
For weak acid HB, ${K_a} = X \times {10^{ - 3}}$ , we have to find out the value of $X$ .
Now when acid HB dissociates-
$HB \to {H^ + } + {{\rm B}^ - }$

Initial	$1$	$0$	$0$
final	$1 - \alpha $	$\alpha $	$\alpha $

Initially only 1 mole of HB was present. After dissociation we get $\alpha $ moles of each ions and $1 - \alpha $ mole of HB.
Now applying the formula
$i = \dfrac{{\pi V}}{{nRT}}$ and $i = 1 + \alpha $
$1 + \alpha = \dfrac{{\pi V}}{{nRT}}$
The relation between number of moles, volume and molarity is given by-
$M = \dfrac{n}{V}$
Where $M$ is molarity and $n$is number of moles.
We have been given that the molarity of acid HB is $0.01M$. So $\dfrac{n}{V} = 0.01$
Substituting the values,
$1 + \alpha = \dfrac{{0.345}}{{0.01 \times 0.082 \times 298}}$
Where $R = 0.082lit - atm mol{e^{ - 1}}{K^{ - 1}}$
$T = 298K$
On calculating the above equation we get the value of $\alpha $ .
$\alpha = 0.411$
Now, the relation between ${K_a}$ and $\alpha$ is given by the formula
${K_a} = \dfrac{{c{\alpha ^2}}}{{1 - \alpha }}$
Where $c$ is concentration . Now substituting the values for acid HB
$X \times {10^{ - 3}} = \dfrac{{0.01 \times {{\left( {0.411} \right)}^2}}}{{1 - 0.411}}$
On solving this equation we get ${K_a} = 2.86 \times {10^{ - 3}}$ for acid HB.
So the nearest integer to $X$ is $3$ .

Note:
-The dissociation constant is dimensionless quantity as it is a constant.
-Molarity, normality and molality are the ways to express concentration of solution.
-With the help of osmotic pressure, molar mass of solute can also be determined.