What is the % ionization of a 0.75 m Na3PO4(aq) solution that has a freezing point of -4.6 0C (kf for water is 1.86 0C/m)?
82.4%
1.6%
96.8%
54.9%
41.2%
To find the % ionization, we need to calculate the freezing point depression caused by the solute and compare it to the expected value for a fully ionized solution.
First, we need to calculate the molality of the solution using the formula:
molality (m) = moles of solute / mass of solvent (kg)
Since we know the molarity of the solution is 0.75 M and the molar mass of Na3PO4 is 163.9 g/mol, we can calculate the number of moles of Na3PO4:
moles of Na3PO4 = molarity x volume in liters = 0.75 M x 1 L = 0.75 mol
Next, we need to calculate the mass of the solvent (water) in kg. Since 1 L of water has a mass of 1 kg, we can assume the mass of solvent is also 1 kg.
Now we can calculate the molality:
m = moles of solute / mass of solvent (kg) = 0.75 mol / 1 kg = 0.75 m
Next, we can use the equation:
∆Tf = Kf x m
Where ∆Tf is the freezing point depression, Kf is the freezing point depression constant for water (1.86 0C/m), and m is the molality of the solution. Rearranging the equation, we can solve for ∆Tf:
∆Tf = Kf x m = 1.86 0C/m x 0.75 m = 1.395 0C
The freezing point of the solution is -4.6 0C, so the expected freezing point of pure water is 0 0C. The freezing point depression is the difference between these two values:
∆Tf_actual = 0 0C - (-4.6 0C) = 4.6 0C
Finally, we can calculate the % ionization using the equation:
% ionization = (∆Tf_actual / ∆Tf_fully ionized) x 100
In a fully ionized solution, Na3PO4 dissociates into 4 ions (3 Na+ ions and 1 PO43- ion). Therefore, the expected freezing point depression for a fully ionized solution is:
∆Tf_fully ionized = 4 x Kf x m = 4 x 1.86 0C/m x 0.75 m = 5.58 0C
Now we can calculate the % ionization:
% ionization = (∆Tf_actual / ∆Tf_fully ionized) x 100 = (4.6 0C / 5.58 0C) x 100 = 82.4%
Therefore, the % ionization of the 0.75 M Na3PO4(aq) solution is 82.4%.