Question

When a decimolar solution of potassium ferricyanide is $50\% $ dissociated at $300{\text{ K}}$. Calculate the osmotic pressure of the solution.

Answer 1

Hint: To solve this we must know that a decimolar solution of potassium ferricyanide means that the concentration of the solution of potassium ferricyanide is $0.1{\text{ M}}$. The formula for potassium ferricyanide is ${{\text{K}}_{\text{4}}}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]$. ${{\text{K}}_{\text{4}}}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]$ dissociates into four potassium i.e ${{\text{K}}^ + }$ cations and one ferricyanide i.e. ${\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^ - }$ anion.

Complete step-by-step answer:
We are given a decimolar solution of potassium ferricyanide. A decimolar solution of potassium ferricyanide means that the concentration of the solution of potassium ferricyanide is $0.1{\text{ M}}$.
The formula for potassium ferricyanide is ${{\text{K}}_{\text{4}}}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]$. ${{\text{K}}_{\text{4}}}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]$ dissociates into four potassium i.e ${{\text{K}}^ + }$ cations and one ferricyanide i.e. ${\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^ - }$ anion. The dissociation reaction of potassium ferricyanide is as follows:
${{\text{K}}_{\text{4}}}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right] \rightleftharpoons 4{{\text{K}}^ + } + {\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^ - }$
At equilibrium, the concentrations of all the species are as follows:
${{\text{K}}_{\text{4}}}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right] \rightleftharpoons 4{{\text{K}}^ + } + {\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^ - }$
$1 - \alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ 4}}\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ }}\alpha $
The total number of moles at equilibrium are:
Total number of moles $ = \left( {1 - \alpha } \right) + 4\alpha + \alpha $
Total number of moles $ = 1 + 4\alpha $

We are given that the solution of potassium ferricyanide is $50\% $ dissociated at $300{\text{ K}}$. Thus, $\alpha = 50\% = 0.5$.
Calculate the osmotic pressure of the solution using the equation as follows:
\[\pi = i \times M \times RT\]
Where, \[\pi \] is the osmotic pressure,
\[M\] is the molar concentration of the solution,
\[R\] is the universal gas constant,
\[T\] is the temperature in kelvin,
\[i\] is the van’t Hoff factor of the solute.
We can calculate the van’t Hoff factor as follows:
\[i = \dfrac{{1 + 4\alpha }}{1} = 1 + 4 \times 0.5 = 3\]
Substitute $0.1{\text{ M}}$ for the molar concentration of the solution, $0.0821{\text{ L atm }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$ for the universal gas constant, $300{\text{ K}}$ for the temperature, ${\text{3}}$ for the van’t Hoff factor. Thus,
\[\Rightarrow \pi = 3 \times 0.1{\text{ M}} \times 0.0821{\text{ L atm }}{{\text{K}}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}} \times 300{\text{ K}}\]
\[\Rightarrow \pi = 7.38{\text{ atm}}\]

Thus, the osmotic pressure of the solution is \[7.38{\text{ atm}}\].

Note: The pressure applied to a pure solvent so that it does not pass into the given solution by osmosis is known as the osmotic pressure. Remember to calculate the van’t Hoff factor correctly. van’t Hoff factor is the number of ions a compound can form when dissolved in water.