Question

Calculate the freezing point of solution containing $8.0g$ of HBr in $100g$ of water assuming the acid to be $90\%$ ionized.

Answer 1

We have to calculate the van‘t Hoff factor from the ionized solution. We have to calculate the freezing point of the solution using the formula of Freezing point depression constant.

Formula used:
The freezing point depression is proportional to the concentration of the solute particles and is given by the equation,
$\Delta {T_f} = im{k_f}$
Here,
$\Delta {T_f} =$Freezing point of the solution
$m =$Molal concentration
${k_f} =$Freezing point depression constant (depends on the solvent used).
$i =$van’t Hoff factor

Complete step by step answer:
Given data contains,
The percent of ionization of solute is $90\%$.
The degree of ionization is calculated as,
$\alpha = \dfrac{{90}}{{100}} = 0.9$
We know that hydrogen bromide dissociates into hydrogen ion and bromide ion. The dissociation reaction is written as,
$HBr\left( {aq} \right) \rightleftarrows {H^ + }\left( {aq} \right) + B{r^ - }\left( {aq} \right)$
We can calculate the van’t Hoff factor from degree of ionization using the formula,
$i = 1 + \alpha \left( {n - 1} \right)$
Here n=number of ions formed from one formula unit
$\alpha$=degree of ionization
We know that for HBr the number of ions formed from one formula unit is two.
The value of degree of ionization is ${\text{0}}{\text{.9}}$.
Substituting these values we can calculate the van’t Hoff factor as,
$i = 1 + \alpha \left( {n - 1} \right)$
Now we can substitute the known values we get,
$i = 1 + 0.9\left( {2 - 1} \right)$
On simplification we get,
$i = 1.9$
The value of van’t Hoff factor is $1.9$
Let us now calculate the freezing point of the solution.
The mass of solute is $8.0g$.
The mass of solvent is $100g$.
Molal depression constant is $1.86Kkg/mol$.
Molar mass of bromine is $80g/mol$.
Molar mass of hydrogen is $1g/mol$.
Molar mass of hydrogen bromide is $81g/mol$.
Initially, let us calculate the change in freezing point of the solution.
$\Delta {T_f} = im{k_f}$
Now we can substitute the known values in the above equation we get,
$\Delta {T_f} = 1.9 \times \dfrac{{8.0g}}{{81g/mol}} \times \dfrac{{1000kg}}{{100g}} \times 1.86Kkg/mol$
On simplification we get,
$\Delta {T_f} = 3.49^\circ C$
We have calculated the change in freezing point of the solution as $3.49^\circ C$.
From the change in freezing point of the solution, we can calculate the freezing point of the solution using the formula,
$\Delta {T_f} = {T_f} - {T_s}$
Now we substitute the values we get,
${T_f} = 0^\circ C - 3.49^\circ C$
On simplification we get,
${T_f} = - 3.49^\circ C$

The freezing point of the solution is $- 3.49^\circ C$.

Note: We must know that the F\freezing point depression is a colligative property. We can also calculate the freezing point depression constant and molality using the expression of freezing point depression of solution.
We can calculate the van’t Hoff factor using the equation,
$i = \dfrac{{{\text{Measured colligative property}}}}{{{\text{Expected value for nonelectrolyte}}}}$
For solutions that are nonelectrolytes, like urea and sucrose, the van’t Hoff factor is one. For solutions of salts and other electrolytes, the value of i is greater than one. For dilute solution such as (${\text{0}}{\text{.01m}}$ or less), the van’t Hoff factor will be equal to the number of ions formed by each formula unit of the compound that dissolves.