Question

Question #0b448

Answer 1

Here's what's going on here.

Explanation:

In order to be able to understand what's going on when the solution is cooled from `60^@"C"` to `0^@"C"`, you need to examine the solubility graph for potassium chloride, `"KCl"`.

Notice that the solubility curve shows that at `60^@"C"`, potassium chloride has a solubility of about `"45 g/100 g H"_2"O"`.

What that means is that at that temperature, you can only dissolve `"45 g"` of potassium chloride per `"100 g"` of water. This exact amount of potassium chloride will result in a saturated solution.

A saturated solution is a solution in which you have an equilibrium between the solid solute and particles of solute that are being dissolved into solution.

At `60^@"C"`, if you add less than `"45 g"` of potassium chloride per `"100 g"` of water you will have an unsaturated solution, i.e. a solution that can still dissolve solute.

If you add more than `"45 g"` of potassium chloride per `"100 g"` of water, you will be left with undissolved solid in solution.

Now, notice that the solubility of potassium chloride changes with temperature. As temperature decreases from `60^@"C"` to `0^@"C"`, the solubility decreases.

This means that the solution will hold increasingly smaller amounts of potassium chloride.

As temperature decreases, the average kinetic energy of the water molecules decreases as well. This implies that the water molecules will be less effective at overcoming the ionic bonds that are holding the solute particles together, and thus less effective at breaking apart the solid's crystal structure.

Moreover, decreasing the temperature of the solution also stabilizes the solid's crystal structure because the vibration of the solid's particles will decrease.

As a result, fewer potassium cations, `"K"^(+)`, and chloride anion, `"Cl"^(-)`, will be hydrolyzed, i.e. split from the crystal structure and solvated by the water molecules.

At `0^@"C"`, the solubility graph shows that potassium chloride has a solubility of about `"28 g/100 g H"_2"O"`.

This tells you that the difference between the amount of potassium chloride that could be dissolved at `60^@"C"` and the amount that could be dissolved at `0^@"C"` will precipitate out of solution.

In simple terms, the solvated ions will once again become part of the crystal structure. Dissolved potassium chloride will become solid potassium chloride. For these values, you will have

`m_"precipitate" = "42 g" - "28 g" = "14 g KCl" ->` will precipitate

`m_"dissolved" = "28 g KCl" ->` will remain dissolved

Sometimes if you cool the solution slow enough, you will get a supersaturated solution, which is a solution that can hold more dissolved solute than the solubility limit at that given temperature.