Answer: The partition coefficient of X between benzene and water is 0.
Explanation: To find the partition coefficient (K) of the tribasic acid (X) between benzene and water, we first need to calculate the amount of X that dissolves in each solvent.
Let's start by finding the amount of X that dissolves in water:
2.8 g of X is shaken with 50 cm^3 of water, which is equivalent to a mass/volume concentration of 0.056 g/cm^3.
Let's assume that x g of X dissolves in 50 cm^3 of water. This means that the concentration of X in water is x/50 g/cm^3.
Since the acid is tribasic, it reacts with sodium hydroxide in a 1:3 stoichiometric ratio. Therefore, the amount of sodium hydroxide needed to neutralize X in water is (1/3) * (x/50) = x/150 moles/cm^3.
We are given that 14.5 cm^3 of 1M NaOH solution is needed to neutralize the X in 25 cm^3 of aqueous layer. This means that the concentration of X in the aqueous layer is (14.5/25) moles/cm^3.
Setting the two expressions for concentration of X equal to each other and solving for x gives:
x/50 = 14.5/25 * 150
x = 261 g
Next, let's find the amount of X that dissolves in benzene:
We are given that X is insoluble in benzene, so the amount of X that dissolves in benzene is effectively zero.
Finally, we can calculate the partition coefficient:
K = (concentration of X in benzene) / (concentration of X in water)
Since the concentration of X in benzene is effectively zero, we have:
K = 0 / (261/50)
K = 0
Therefore, the partition coefficient of X between benzene and water is 0.