0.1M weak acid at 27 degree celcius has osmotic pressure of 2.5 atm. find % dissociation of acid?
To find the percent dissociation of a weak acid, we need to consider the concept of osmotic pressure.
Osmotic pressure is a colligative property that depends on the total number of particles present in a solution. For dilute solutions, where the solute is non-volatile and does not dissociate completely, we can use the Van't Hoff factor (i) to determine the effective number of particles.
In the case of a weak acid, it partially dissociates into ions in water, so we need to consider the degree of dissociation (α). The formula for osmotic pressure (π) is given as:
π = i * M * R * T
Where:
π = Osmotic pressure
i = Van't Hoff factor (effective number of particles)
M = Molarity of the solution
R = Ideal gas constant (0.0821 L*atm/(mol*K))
T = Temperature in Kelvin
Given:
Molarity (M) = 0.1M
Osmotic pressure (π) = 2.5 atm
Temperature (T) = 27°C = 273 + 27 = 300K
Assume i = 2 for the dissociation of a weak acid.
Rearranging the formula, we can solve for α (percent dissociation):
α = (π / (i * M * R * T)) * 100
Now, let's substitute the values into the equation and calculate α:
α = (2.5 / (2 * 0.1 * 0.0821 * 300)) * 100
Calculating the above expression, we find:
α ≈ 0.101 ≈ 10.1%
Therefore, the percent dissociation of the weak acid at 27 degrees Celsius is approximately 10.1%.