a decimolar solution of potassium ferrocyanide is 5% dissociated at 300 Kelvin calculate Osmotic pressure of solution
When the complex is not dissociated then we can use the Van 't Hoff equation to calculate the normal osmotic pressure.
P = c*R*T
Where, P = Osmotic pressure,
c = concentration
R = gas constant
T = temperature in Kelvin
Given values, c= 0.1 molar , T = 300 K
P = 0.1 * 0.0821 * 300
P = 2.462 atm
However, the given complex is ionisable and it dissociates as follows....
K4[Fe(CN)6] → 4K+ + [Fe(CN)6]4-
1-a 4a a
where, a is the degree of dissociation of the complex.
Initial moles of complex = 1
Moles after dissociation of the complex= 1 - a + 4a +a = 1 + 4a
Since, osmotic pressure is directly proportional to the number of moles, hence
P(observed)/P(normal) = (1+4a)/1
Dissociation takes place only 5%. so, a = 0.050
P/2 x 462 = 1+(4*0.050)
P/2 x 462 = 1 + 0.2
P/2 x 462 = 1.2
P = 2.462 x 1.2
P = 2.9544 atm
When the complex is not dissociated then we can use the Van 't Hoff equation to calculate the normal osmotic pressure.
P = c*R*T
Where, P = Osmotic pressure,
c = concentration
R = gas constant
T = temperature in Kelvin
Given values, c= 0.1 molar , T = 300 K
P = 0.1 * 0.0821 * 300
P = 2.462 atm
However, the given complex is ionisable and it dissociates as follows....
K4[Fe(CN)6] → 4K+ + [Fe(CN)6]4-
1-a 4a a
where, a is the degree of dissociation of the complex.
Initial moles of complex = 1
Moles after dissociation of the complex= 1 - a + 4a +a = 1 + 4a
Since, osmotic pressure is directly proportional to the number of moles, hence
P(observed)/P(normal) = (1+4a)/1
Dissociation takes place only 5%. so, a = 0.050
P/2 x 462 = 1+(4*0.050)
P/2 x 462 = 1 + 0.2
P/2 x 462 = 1.2
P = 2.462 x 1.2
P = 2.9544 atm
