Question

For an aqueous solution of hf, determine the van't hoff factor assuming 0% and 100% ionization, respectively. a solution is made by dissolving 0.0100 mol hf in enough water to make 1.00 l of solution. at 22 °c, the osmotic pressure of the solution is 0.307 atm. what is the percent ionization of this acid?

Answer 1

The Van't Hoff factor represents the degree of ionization in a solution. For HF, with 0% ionization, i=1 and with 100% ionization, i=2. The actual ionization degree can be found by measuring the solution's osmotic pressure and using it to solve for i.

Step-by-step explanation:

The Van't Hoff factor (i) is used to understand the degree of ionization in a solution. When considering hydrofluoric acid (HF) in an aqueous solution, we can determine two ideal van't Hoff factors: one assuming 0% ionization and another assuming 100% ionization. At 0% ionization, HF remains as intact molecules, so i = 1. Fully ionized at 100%, HF dissociates into H+ and F- ions, which makes i = 2.

Given the osmotic pressure and molarity of the solution, we can calculate the actual degree of ionization. The osmotic pressure (Π) follows the equation Π = iMRT, where M is the molarity, R is the ideal gas constant, and T is the temperature in Kelvin. By using the actual osmotic pressure measured (0.307 atm) for a 0.0100 M HF solution and solving for i, we can then compare this to the ideal factor to determine the percent ionization of HF.

Answer 2

According to Osmotic pressure equation:

π = i M R T

When π =0.307 atm & M = 0.01 mol & R (constant)= 0.0821 L-atom/mol-K &
T= 22+273 = 295 Kelvin

So Van't half vector i = π / (MRT)
= 0.307 / (0.01 * 0.0821 * 295)
= 1.27

When there is no dissociation, i = no. of moles of Hf in 1 L of solution = (1-X)
and when there is a complete dissociation so it is equal 2X according to this equation


HF(aq) + H2O (L) ⇆ H3O (aq) + F (aq)
(1-X) X X

∴ i = (1-X) + (2x)

1.27 = 1+X
∴X= 1.27 - 1 = 0.27
∴ the percent ionization of the acid X = 27 %