Question

Given,15.0g of a monobasic acid, when dissolved in 150g of water, lowers the freezing point by ${{0.165}^{0}}C$. 0.5g of the same acid when titrated, after dissolution in water, requires 37.5ml of N/10 alkali. Calculate the degree of dissociation of the acid? ( ${{K}_{f}}$ for water = $1.86{{\,}^{0}}C\,mo{{l}^{-1}}$ ).

Answer 1

Hint: The concept of the depression in the freezing point where the formula for its calculation is given by, $\Delta {{T}_{f}}={{K}_{f}}\times m\times i$ and by finding the molality ‘m’ and also of ‘i’ which is the Van’t Hoff factor, you will approach the required answer

Complete answer:
We have studied the freezing point depression and the related calculation of degree of dissociation of acid and many other concepts related to it.
Let us now calculate the degree of dissociation by using the given data.
- Depression in freezing point is the decrease in the freezing point of the solvent when the non volatile solute is added to it.
- The formula to find the depression in the freezing point is given by,
$\Delta {{T}_{f}}={{K}_{f}}\times m\times i$ ……..(1)
where, $\Delta {{T}_{f}}$ is the freezing point depression.
${{K}_{f}}$ is the cryoscopic constant.
$m$is the molality of the solution and
$i$ is the Van’t Hoff factor.
This Van’t Hoff factor is different for different solutes. It defines the extent of the association or dissociation of solutes in solution.
Now, in the above given data, we have $\Delta {{T}_{f}}={{0.165}^{0}}C$, ${{K}_{f}}=1.86{{\,}^{0}}C\,mo{{l}^{-1}}$
Now, molality of the solution will be, $m=\dfrac{{{n}_{solute}}}{Weight\,of\,solvent(kg)}$
The number of moles of solute will be its mass divided by the molar mass and hence the above equation will be,
$m=\dfrac{1.5}{M\times 150}\times 1000=\dfrac{10}{M}$
Van’t Hoff factor $i=1+(n-1)\alpha $
For monobasic acid, the value of n will be 2 and hence, $i=1+(2-1)\alpha =1+\alpha $
Now, substituting all these values in equation (1), we have
$0.165=1.86\times \dfrac{10}{M}\times (1+\alpha )$ …………(2)
Now, since it is given in the data as, 0.5g of the same acid when titrated, after dissolution in water, requires 37.5ml of N/10 alkali, we can find the molar mass of the compound that is,
$\dfrac{0.5}{M}=37.5\times {{10}^{-3}}$
Thus, $M=\dfrac{0.5}{37.5\times {{10}^{-3}}}=133.33g/mol$
Substituting this in equation (2), we get
$0.165=1.86\times \dfrac{10}{133.33}\times (1+\alpha )$
By, solving for $\alpha $ , we get $\alpha =0.1827$
To convert this into percentage, we will multiply this value by 100 and that will be, $\alpha =0.1827\times 100=18.27$ %

Note:
Note that monobasic acid dissociates into single cation and single anion and thus there are total of two ions which means that the value of ‘n’ will be 2 which makes the value of the Van’t Hoff factor for monobasic acid as, $1+\alpha $.