# Question #71458

Jul 23, 2015

Simply put, you dissolve a sample of solid potassium iodide in water.

#### Explanation:

Since you didn't give any detail about your target solution, I'll show you how to make an unsaturated, a saturated, and a supersaturated solution of potassium iodide.

Here's how potassium iodide's solubility graph looks like

So, let's say that you want to make an unsaturated potassium iodide solution at ${20}^{\circ} \text{C}$. According to the solubility graph, you cannot dissolve more than approximately 160 g of potassium iodide per 100 mL of water.

This means that, in order to make an unsaturated solution, you need to dissolve less than 160 g of potassium iodide per 100 mL of water.

To make a saturated solution, you need to add enough potassium iodide per 100 mL of water to match the solubility at that respective temperature.

In this case, if you slowly dissolve 160 g of potassium iodide per 100 mL of water at ${20}^{\circ} \text{C}$, you'll make a saturated solution.

To get a supersaturated solution, you need to find a way to increase the amount of solvent that can be dissolved per 100 mL in your solution.

Let's say that you want your solution to dissolve as much as 180 g of potassium iodide per 100 mL of awter at ${20}^{\circ} \text{C}$.

To do that, you heat the saturated solution to, say, ${40}^{\circ} \text{C}$. At this temperature, the solution can hold a maximum of approximately 190 g of potassium iodide per 100 mL of water.

At ${40}^{\circ} \text{C}$, 180 g of potassium iodide will actually make for an unsaturated solution, since the maximum solubility at this temperature is 190 g per 100 mL.

After you dissolve 180 g of potassium iodide in the solution, you slowly cool the solution down to ${20}^{\circ} \text{C}$.

If you manage to do this correctly, the solution will now hold 180 g of potassium iodide per 100 mL of water at ${20}^{\circ} \text{C}$, i.e. it will be spersaturated.