A solution is made by dissolving 0.0150 mol of HF in enough water to make 1.00 L of solution. At 26 °C, the osmotic pressure of the solution is 0.449 atm. What is the percent ionization of this acid?

Question

anonymous · Answer

so 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%