A solution is made by dissolving 0.0100 mol of HF in enough water to make 1.00 L of solution. At 23 °C, the osmotic pressure of the solution is 0.308 atm. What is the percent ionization of this acid?
Please help.
I have seen another problem and I doubted it completely. This one I don't believe but at least it's possible. I will assume the author is just making up problems.
pi = iMRT
Plug in all the numbers ans solve for i = van't Hoff factor.
Then i*0.01 = concn the 0.01 appears. I think it's about 0.013 but you need to clean up the numbers.
The ionization of HF is
.............HF == H^+ + F^-
initial....0.01....x.....x
change......-x.....+x....+x
equil.....0.01-x....x......x
So what do we have in solution?
We have x + x + 0.01-x = ? and it APPEARS to be 0.013
Solve for x which will give you the ion concn, then %ionization = 100*(x/(0.01) = ?
To find the percent ionization of the acid (HF) in the given solution, we need to calculate the concentration of the ions and the initial concentration of the acid. Then, we can use the formula for percent ionization.
First, let's calculate the initial concentration of HF:
We know that 0.0100 mol of HF is dissolved in 1.00 L of solution.
Therefore, the initial molar concentration of HF is:
concentration = moles/volume
concentration = 0.0100 mol / 1.00 L
concentration = 0.0100 M (M stands for molar)
Next, let's calculate the concentration of H+ ions after ionization:
For every 1 molecule of HF that ionizes, it produces 1 H+ ion and 1 F- ion. This means that the concentration of H+ ions will be the same as the concentration of HF that ionized.
Assuming x is the amount of HF that ionizes, the concentration of H+ ions can be given as:
[H+] = x
Since we started with 0.0100 M concentration of HF, the concentration of the remaining un-ionized HF would be:
[HF] remaining = 0.0100 M - x
Now, we can use the formula for osmotic pressure to relate ion concentration to osmotic pressure:
π = MRT
Where:
π = osmotic pressure
M = molar concentration
R = ideal gas constant (0.0821 atm L/mol K)
T = temperature (in K)
Rearranging the formula to solve for M:
M = π / RT
Plugging in the given values:
M = 0.308 atm / (0.0821 L atm/mol K * 23 + 273 K)
M ≈ 0.308 atm / 24.0 L atm/mol
M ≈ 0.0128 M
We substitute the value of the molar concentration of H+ ions ([H+]) into the equation for the percent ionization:
% ionization = ([H+] / initial concentration of HF) × 100
% ionization = (x / 0.0100) × 100
To find the value of x, we can use the equation:
x = 0.0128 M (from the calculated concentration of H+ ions)
Substituting this back into the equation for % ionization:
% ionization = (0.0128 / 0.0100) × 100
% ionization ≈ 128
Therefore, the percent ionization of the acid (HF) in the solution is approximately 128%.