Ethoxyethane, C₂H₅OC₂H₅, can dissolve both in water and in octan-1-ol. The expression and numerical value for the partition coefficient of ethoxyethane between water and octan-1-ol are given. Water and octan-1-ol are immiscible. Kpc = concentration of C₂H₅OC₂H₅ in octan-1-ol / concentration of C₂H₅OC₂H₅ in water = 6.760 at 20°C

1. (a)In an experiment, octan-1-ol at 20°C is added to a solution of ethoxyethane in water at 20°C. The mixture is analysed immediately and a value of Kpc is calculated. The calculation is performed correctly; the value calculated is 5.625. Explain why the value calculated is less than 6.760.[2]
2. (b)(i)A second experiment is performed and the value of Kpc is found to be 6.760. The concentration of ethoxyethane in the octan-1-ol layer is 7.62g dm⁻³. Calculate the concentration, in gdm⁻³, of ethoxyethane in the aqueous layer.[1]
3. (b)(ii)100 cm³ of the octan-1-ol layer is taken and shaken with 100cm³ of water. Calculate the maximum amount, in mol, of ethoxyethane that can be extracted into the water.[3]
4. (c)(i)An aqueous solution of lead(II) nitrate is mixed with an aqueous solution of sodium iodide. A yellow precipitate of lead(II) iodide is formed and is filtered out, leaving solution X. The concentration of Pb²⁺ in solution X is 5.68 × 10⁻³ mol dm⁻³. The concentration of I⁻ in solution X is 4.20 × 10⁻⁴ mol dm⁻³. Use these data to calculate a value for the solubility product, Ksp, of lead(II) iodide. State the units of Ksp.[2]
5. (c)(ii)Potassium iodide is very soluble in water. Describe and explain what is seen if a few drops of saturated potassium iodide solution are added to a portion of solution X.[2]