Question

100 ml of 0.1 M solution of a weak acid HA has molar conductivity 5$Sc{m^2}mo{l^{ - 1}}$. The osmotic pressure of resulting solution obtained after dilution of original solution upto 1 litre at 500 K assuming ideal solution is
$\left( {Given:\lambda _m^\infty ({H^ + }) = 450Sc{m^2}mo{l^{ - 1}},\lambda _m^\infty ({A^ - }) = 50Sc{m^2}mo{l^{ - 1}},R = 0.08lt - atm/mol - K,\sqrt {10} = 3.2} \right)$.
Report your answer as X, where X= Osmotic pressure (in atom)$\times$10. Give your answer at the nearest integer.

Answer 1

Weak acid does not completely dissociate in the solution which causes error in the answer. It means that when a weak acid is added to a solution it does not completely furnish all its hydrogen ions into the solution. So here we have to consider the extent of dissociation here.

Complete step by step answer:
In this equation, we have to find the value of osmotic pressure. Osmotic pressure is the minimum excess pressure that has to be applied on the solution to prevent the flow of the solvent into it through the semipermeable membrane. It is denoted by $'\pi '$. Moreover we have to understand about molar conductivity and limiting molar conductivity. The conductance of all the ions produced by dissolving one mole of electrolyte in solution is called molar conductivity. It is denoted by ${\lambda _m}$. The molar conductivity at infinite dilution is called limiting molar conductivity and is denoted by $\lambda _m^\infty$ (or $\lambda _m^o$)
In the question, we have given that 100 ml weak acid has molarity 0.1M.
As we know that, molarity =$\dfrac{{Number{\text{ of moles of solute }} \times {\text{1000}}}}{{Volume{\text{ of solution(in ml)}}}}$
Putting values in above equation, we obtain
0.1= $\dfrac{{Number{\text{ of moles of solute}} \times {\text{1000}}}}{{100ml}}$
Hence the number of moles of solute is $\dfrac{{0.1}}{{10}}$=0.01 mol -------(1)
Also we have given molar conductivity be 5 $Sc{m^2}mo{l^{ - 1}}$
Now, we have to find limiting molar conductivity of weak acid HA as $\lambda _m^\infty ({H^ + }) = 450Sc{m^2}mo{l^{ - 1}}$ and $\lambda _m^\infty ({A^0}) = 50Sc{m^2}mo{l^{ - 1}}$, where $\lambda _m^\infty ({A^0})$ is limiting molar conductivity = $\lambda _m^\infty ({H^ + }) + \lambda _m^\infty ({A^0})$ of HA will be $\left[ {\lambda _m^\infty (HA)} \right]$
Or $\lambda _m^\infty (HA) = 450Sc{m^2}mo{l^{ - 1}} + 50Sc{m^2}mo{l^{ - 1}}$
$\lambda _m^\infty (HA) = 500Sc{m^2}mo{l^{ - 1}}$------------(2)
As we know that degree of dissociation is expressed as molar conductivity or solute divided by limiting molar conductivity of solute. So this implies that $\alpha = \dfrac{{{\lambda _m}}}{{\lambda _m^\infty (HA)}}$
Putting the values of ${\lambda _m}$ and $\lambda _m^\infty (HA)$, we get
$\alpha = \dfrac{{5Sc{m^2}mo{l^{ - 1}}}}{{500Sc{m^2}mo{l^{ - 1}}}} = 0.01$-------------(3)
Since the given acid is weak, therefore we have to consider the van't Hoff factor here. It expresses the extent of dissociation or association of solute and solution. We have given acid, so obviously dissociation takes place. So we calculate went van't hoff factor (i) by using following expression:-
i.e. $i = 1 - \alpha + n\alpha$-------------(4)
Here n is the number of particles after dissociation. Since HA dissociates into 2 particles ${H^ + }{\text{ and }}{{\text{A}}^ - }$, so here n becomes equal to 2. Moreover after putting values of$\alpha$ from equation (3) in equation (4), we get:-
i= 1-0.01+2(0.01)
i= 1.01-----------(5)
Moreover the solution is diluted upto 1 litre solution, so new molarity becomes
Molarity(c)= $\dfrac{{Number{\text{ of moles of solute}}}}{{Volume{\text{ of solution(in l)}}}}$
Putting the values from equation (1) in above equation, we get
C=$\dfrac{{0.01}}{1} = 0.01M$-----------(6)
As we know we calculate osmotic pressure using formula $\pi = i.CRT$ here C is molarity, R is ideal gas constant and T is temperature. Putting values from equation (6) and (5) in above equation, we get
$\pi = 1.01 \times 0.01M \times 0.08L{\text{ at m mo}}{{\text{l}}^{ - 1}} \times 500K$
On calculating, we get
$\pi = 0.404atm = osmotic{\text{ pressure}}$
Now we have to find X= osmotic pressure$\times 10$
= 0.404$\times 10 = 4.04$
Since we have to give an answer in integer, the nearest integer is 4. Hence the answer is ‘4’.

Note:
We use the van't Hoff factor in this equation because of weak acid. Weak acid does not dissociate properly. So to overcome the error in the answer, we use the vent Hoff factor.