Question

A solution prepared by dissolving 10 grams of HF in 500 grams of water, freezes at a temperature of -1.98 ° C. Calculate the pH of this solution and the dissolution constant of the HF acid solution with density d = 1.05 g / mL. (Freezing constant = - 1.86 ° C / m)

Answer 1

I think you must have meant dissociation constant otherwise known as the ionization constant (Ka) instead of dissolution constant.
molality = mols/kg solvent and mols = grams/molar mass.
molar mass HF = 19 + 1 = 20 grams/mol in the problem = 10 g so
mols = 10/20 = 0.5 mols. The problem lists 500 mL or 0.5 kg as the solvent. Then m = mol/kg solvent = 0.5mol/0.5 kg = 1 molal = 1 m.

delta T = i*Kf*m
delta T = 1.98; Kf = 1.86; m = 1 so
i = 1.98/1.86*1 = 1.064
Solution is 1 m but it acts like 1.064 m, the difference is
1.064-1.000 = 0.064
So % ion = 0.064/1.000)*100 = 6.4% (I don't believe this but that's what the numbers say.)
Using the density to convert molality, m, to molarity, M.
mass = volume x density. You know the density in the problem is 1.05 g/mL. You know the water is 500 mL or 500 g and you know the solute is 10 g so total mass is 510 g so volume = mass/density = 510/1.05 = 486 mL or 0.486 L so M = mols/L = 0.5/0.486 L = 1.03 M
.....................HF ==> H^+ + F^-
I...................1.03........0.......0
C....................-x..........x........x
E...............1.03-x.........x........x
If the solution is 6.4% ionized then you know that x = 1.03 * 0.064 = 0.066 then the E line reads as
E...............0.964.........0.066......0.066 then Ka becomes
Ka = (H^+)(F^-)/(HF) = ? Substitute the last E line and solve for Ka.
The pH of the solution is obtained from pH = -log (H^+) and the H^+ you read from above (0.066 M).
Post your work if you get stuck.