A solution is made by dissolving 0.0450 mol of HF in enough water to make 1.00 L of solution. At 25 °C, the osmotic pressure of the solution is 1.24 atm. What is the percent ionization of this acid?

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anonymous · Answer

osmotic pressure = (molarity)(gas constant)(temp) we can use the given information to find the molarity of the solution that would give us that osmotic pressure... so... 0.449atm = (Molarity)(0.08206 Latm/molK)(299K) Molarity = 0.0183 so the total particles in the solution will be this molarity - the given molarity so... 0.0183-0.0150 = 0.0030 M we then put this value over the value given molarity to determine the percent ionization. 0.0030M/0.0150M = 0.200 = 20.0% http://openstudy.com/study#/updates/4f413614e4b0534c53cc568e