Question

A solution is made by dissolving 0.0500 mol of HF in 1.00 kg of water. The solution was found to freeze at –0.104 °C. Calculate the value of i and estimate the percent ionization of HF in this solution.

Answer 1

Answer: i = 1.12 and percent ionization is 12% .

Explanation: 0.0500 moles of HF are dissolved in 1.00 kg of water. Molality of solution is moles of solute divided by the kg of solvent. Since the mass of solvent is just 1.00 kg, the molality of the solution will be 0.0500m.

The formula used for depression in freezing point is written as:

$\Delta T_f=imk_f$

where, i stands for Van't hoff factor, m stands for molality of solution and kf stands for molal freezing point constant for the solvent.

Pure water freezes at zero degree C and the freezing point of the solution is given as -0.104 degree C. It means, $\Delta T$ =0.104 degree C

kf for water is 1.86 degree C/m .

Let's plug in the values in the formula and calculate the value of i.

$0.104=i*0.0500*1.86$

$0.104=i*0.093$

$i=\frac{0.104}{0.093}$

i = 1.12

Van't hoff factor and degree of ionization are related to each other as:

$i=1+\alpha (n-1)$

where alpha stands for degree of ionization and n is the number of ions formed from one formula unit of the substance.

The equation for the ionization of HF is:

$HF\rightarrow H^++F^-$

from above equation, the value of n is 2 as two ions are formed.

Let's plug in the values in the formula and calculate the value of alpha.

$1.12=1+\alpha (2-1)$

$1.12=1+\alpha (1)$

$1.12=1+\alpha$

$\alpha =1.12-1$

$\alpha =0.12$

Value of alpha is 0.12 means the percent ionization of HF in the given solution is 12% .