A decimolar solution of potassium ferricyanide (K_{4}[Fe(CN)_{6}]) is 50% dissociated at 300k. calculate the 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....

K_{4}[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 50%. so, a = 0.50

P/2 x 462 = 1+(4*0.50)

P/2 x 462 = 1 + 2

P/2 x 462 = 3

P = 2.462 x 3

P = 7.386 atm

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