A solution is made by dissolving 0.500 mol of HF in enough water to make 1.00 L of solution. At 21 C, the osmotic pressure of the solution is 1.35 atm. What is the percent ionization of this acid?
This is what i did.. is it correct?
(1.35 atm) = (2)(M)(0.08206)(294 K)
M = 0.0279
0.0279/0.500 = 0.0559 X 100 = 5.59%? is that correct...? it doesn't seem right please help!
Eric, will you please look at the problem to be sure you have posted it correctly. I have struggled with this because the numbers didn't look right to me and finally figured out why I had doubts. IF (capitalized IF) HF did not ionize at all, then i would be 1. But if we try to calculate i we get
1.35 = i*0.5*0.08206*294
i = about 0.1 and that just can't be. It can't be less than 1 give or take a little to make up for non-ideal solutions and that kind of thing. A 0.05M soln would come closer
I'm sorry! I did post it incorrectly! it was 0.05 M of solution not 0.500!
so would it be.. 1.35 = 1.206i
i = 1.119?
is that when you plug it in with osmotic pressure = i(m)(r)(t)?
i'm sorry for my mistake!
Please be careful. I spent about an hour trying to figure out what was wrong before I suddenly realized that the post had to be wrong.
Yes, solve for i and I get 1.119 also, then 1.119 x 0.05 = about 0.0560 (so it is 0.05 but appears to be 0.0560). A few of the HF molecules break up but only some of them.
.............HF ==>H^+ + F^-
initial.....0.05...0.....0
change.......-x....x.....x
equil......0.05-x...x.....x
We add all of the ions and make it add up to 0.0560
x + x+ 0.05-x = 0.0560
Solve for x = about 0.006
Then %ion = (0.006/0.05)*100 = ?
I am very sorry Dr. Bob =/ thank you so much for the help though I understand now
Your approach is almost correct, but there is a small mistake in your calculation. Let's go through the process step by step to find the correct answer.
The first step is to use the ideal gas law for osmotic pressure, which is derived from the equation:
π = MRT
where:
π is the osmotic pressure,
M is the molarity of the solution (in mol/L),
R is the ideal gas constant (0.08206 L·atm/(mol·K)),
T is the temperature in Kelvin.
Given:
M = 0.500 mol/1.00 L = 0.500 mol/L
π = 1.35 atm
T = 21°C = 21 + 273 = 294 K
Rearranging the equation, we have:
M = π / (RT)
Now we substitute the given values:
M = (1.35 atm) / ((0.08206 L·atm/(mol·K))(294 K))
M ≈ 0.0642 mol/L
So the molarity of the solution is approximately 0.0642 mol/L.
Next, to find the percent ionization, we need to know the initial molarity of the acid and the molarity of the ionized acid. From the balanced chemical equation of HF:
HF ⇌ H⁺ + F⁻
We observe that, for every 1 mol of HF that undergoes ionization, we get 1 mol of H⁺ and 1 mol of F⁻ ions.
Since HF is a monoprotic acid, the initial molarity of HF is equal to the molarity of the acid before any ionization, which is 0.500 mol/L.
The molarity of the ionized acid is then equal to the concentration of H⁺ ions, which is also 0.500 mol/L.
Percent ionization = (Molarity of ionized acid / Initial molarity of acid) × 100
Percent ionization = (0.500 mol/L / 0.500 mol/L) × 100
Percent ionization = 100%
Therefore, the percent ionization of this acid is 100%.
Your calculation using the ideal gas law for osmotic pressure was correct, but you made a mistake in the subsequent calculation to find the percent ionization. The percent ionization of a completely ionized acid is always 100%.