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 â°c, the osmotic pressure of the solution is 0.307 atm. what is the percent ionization of this acid?

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 %

Tolerance to high levels of salinity