Question

There is $KI$ and sucrose with \[0.1{{ }}M\;\] concentration, if the osmotic pressure of $KI$ and sucrose solution is \[0.465{{ }}atm\] and \[0.245{{ }}atm\] respectively. Then find its degree of dissociation for $KI$ .

Answer 1

Hint: Degree of dissociation can be found by finding the Van't Hoff factor $i$ . The Van't Hoff factor can be found using the relation that it has with a colligative property like osmotic pressure as the Van't Hoff factor is the measure of the change that occurs in the solution due to the addition of the solute.

Formula used: osmotic pressure, $\pi = iCRT$$atm$
Where $C$ is the concentration of the solution in the question, $R$ is the gas constant and $T$ is the temperature of the solution. $i$ is the Van't Hoff factor.
Degree of dissociation, $\alpha = \dfrac{{i - 1}}{{n - 1}}$
Where $i$ is the Van't Hoff factor, $n$ is the number of ions formed.

Complete step by step answer:
Osmotic pressure is a colligative property meaning it is dependent on the ratio of the number of the solute particles present in the solution to the solvent molecules present in the solution irrespective of the chemical properties of the constituent particles in the solution.
This also means that it is also related to the Osmotic pressure as the formula mentioned above.
The answer can be obtained using the following steps. It is first required to find the $i$ of the solution.
We have two osmotic pressures here. The van't hoff factor for compounds that are not salts like sucrose is usually $1$ . It is now only required to find the van't Hoff factor of $KI$ . This can be done by the following steps.
The osmotic pressure for $KI$ = $0.465{{ }}atm$
The osmotic pressure for sucrose = $0.245{{ }}atm$
Substituting these values, we get,

$\dfrac{{{\pi _{KI}}}}{{{\pi _{sucrose}}}}$$ = \dfrac{{{i_{KI}}CRT}}{{{i_{sucrose}}CRT}}$
$\dfrac{{0.465{{ }}atm}}{{0.245{{ }}atm}} = \dfrac{i}{1}$
dividing the numerator and denominator, we get,
$1.89 = i$
After finding the $i$ , we can find the dissociation constant, $\alpha $ . Remember that the $n$ of $KI$ is $2$ .
This value of $n$ represents the number of ions that can be formed from the compound $KI$ . since it can form two ions, we consider $n$ to be $2$ . This is carried out in the equations below:
\[\alpha = \dfrac{{i - 1}}{{n - 1}}\]
\[\alpha = \dfrac{{1.89 - 1}}{{2 - 1}}\]
Solving the numerator and denominator we get,
\[\alpha = \dfrac{{0.89}}{1}\]
\[\alpha = 0.89\]

The dissociation constant will be \[0.89\] . The percentage ionization will be $89\% $ .

Note: The $C$ mentioned in the reaction of osmotic pressure is actually molarity of the solution and can also be represented by $\dfrac{n}{V}$ where $n$ is the number of moles and $V$ is the volume of the solution.
The Van’t Hoff factor is also related to all colligative properties as, $i = \dfrac{{observed\;property}}{{theoretical\;property}}$.