Chapter 16, Problem 52A

Interpretation Introduction

Interpretation: The reason is to be explained whether a solution with undissolved solute can be supersaturated or not.

Concept Introduction:

For understanding supersaturated solutions, one should need first understand the different types of solutions:

A solution that can dissolve more solute is said to be unsaturated. There are no visible lumps of undissolved solute in such a solution since all of the solutes are completely dissolved.

Any more solute will now be visible in the solvent as chunks of undissolved particles because a saturated solution is no longer capable of dissolving it. Similar to an unsaturated solution, a saturated one has no undissolved particles; but, at this stage, no additional solute can be dissolved. Between the solute's dissolution and its re-crystallization, there occurs a dynamic equilibrium.

A saturated solution can be made supersaturated by adding additional solute. Undissolved solute particles will now be apparent in the solution. When a solution contains more solute than it is capable of dissolving at a specific temperature and volume, the solution is said to be supersaturated.

Answer to Problem 52A

Supersaturated solutions are any liquids that contain more solute than is necessary to generate a saturated solution at any given constant temperature.

The undissolved solute cannot be present in a supersaturated solution since doing so would reveal that it is a saturated solution.

Explanation of Solution

A supersaturated solution is in a meta-stable state, meaning that as long as it is not disturbed, it will stay that way. It will become a saturated solution in the presence of a small perturbation (such as a slight temperature rise or fall).

For example:

Assume the substance, $X$ dissolves in water at $20°\text{C}$ at a rate of $10\text{g}$ per $100\text{g}$ . Only $10\text{g}$ of solid $X$ will dissolve when added to $100\text{g}$ of water at $20°\text{C}$ and stirred; the remaining $5\text{g}$ will remain undissolved at the bottom of the container. Because at this precise temperature, $10\text{g}$ of $X$ has dissolved, this solution is saturated. This solution becomes unsaturated if you heat it to a higher temperature since warmer solutions can usually dissolve more solute. Therefore, until the solution is once again saturated at a higher temperature, some of the $5\text{g}$ of undissolved solute will dissolve.

The solute will all dissolve if the solution is heated to a high enough temperature, but the solution will still be saturated or unsaturated. When all $15\text{g}$ of $X$ have been dissolved and you let this heated solution gently and carefully cool back to $20°\text{C}$ , you have created a supersaturated solution since more than the maximum mass of $X$ is dissolved at the starting temperature.

As a result, a supersaturated solution is unstable, and even a small physical disturbance might result in the crystallization of the extra dissolved solute $\left(5\text{g}\right)$ .

Conclusion

A supersaturated solution cannot contain undissolved solids in it. For making a supersaturated solution, the solute should be dissolved at a particular undisturbed state.