### Video Transcript

A saturated solution of potassium chloride is formed by addition of the solid to 50 millilitres of water. The solution is filtered and the remaining liquid evaporated. Using the data below, calculate the solubility of potassium chloride at 15 degrees Celsius in grams of salt per 100 mL of water. Use a value of 0.999 grams per milliliter for the density of water at this temperature.

A saturated solution is one that contains as much of the specified material as possible. Here, the material is potassium chloride, KCl. And it’s being added to 50 millilitres of water. Potassium chloride should have a decent solubility in water, given that it’s made of singly charged ions. Excess potassium chloride is removed from the solution by filtration. And then all the water is removed. So we should end up with pure potassium chloride again. We can use the data in the table to figure out the solubility of potassium chloride.

The experiment is being done at 15 degrees Celsius. Our evaporation basin has a mass of 65.32 grams. Before we evaporate away the solvent, the mass of the basin plus the salt solution is 125.32 grams. And the mass of the evaporating basin with our dry potassium chloride is 81.32 grams. Our job is to use this information to calculate the solubility of potassium chloride at 15 degrees Celsius in grams of salt per 100 mL of water.

The solubility of a substance is a measure of its ability to dissolve. The units of solubility are commonly grams per 100 millilitres of solvent. Solubilities can change with temperature. So it’s often important to state at what temperature the measurement is made. The value for the solubility is the number of grams that would dissolve in 100 mL of liquid. Because of the addition of solid, the final volume of solution may actually be greater than 100 millilitres . So for this question, we’re going to need to be extremely careful about what we’re referring to when we’re referring to a volume or to a mass.

In this experiment, we’re starting with 50 millilitres of water. In the next step, we’re adding solid potassium chloride to the liquid. Some of the potassium chloride dissolves, forming a saturated solution. But some solid is left behind. After thorough mixing, the saturated solution is filtered, removing the solid potassium chloride left over. The mass of an evaporating basin is measured. The saturated potassium chloride solution is poured into the basin. And then the total mass is measured. The liquid in the solution is then evaporated over a Bunsen burner. And finally, the mass of the salt plus that of the evaporating basin is measured. Using all this information, we should be able to figure out the mass of potassium chloride that dissolved in our liquid.

Let’s start by working out the final mass of potassium chloride. We can calculate this by taking the total mass of the basin and the salt and taking away the mass of the basin. This is equal to 81.32 grams minus 65.32 grams, which is equal to 16 grams. So we figured out our mass of salt. At this point, it might be tempting to take the mass of salt we’ve worked out and divide it by 50 millilitres from the start of the question. However, some of that water would’ve been lost during the filtration and wouldn’t have made it in to the evaporating basin. So to work out the solubility, what we need to do is work out the mass of water that was in the solution when we poured it into the basin.

To figure out the mass of water in our salt solution, we need to take the mass of the basin plus the mass of the salt solution. Take away the mass of the basin and take away the mass of the salt. This is equal to 125.32 grams minus 65.32 grams minus 16 grams. This gives us a mass of water of 44.00 grams. Now that we’ve worked out the mass of water, we need to work out the volume of this water if it were pure. We’ve been given a density of water at 15 degrees Celsius, slightly lower than the one gram per mL we’re used to. The density of a substance is equal to its mass divided by its volume. So the volume is equal to the mass divided by the density. Alternatively, we can take the mass and multiply it by one mL per 0.999 grams. This gives us a volume of pure water of 44.0440 millilitres.

What we don’t have here is the volume of solution in our basin. Instead, what we have is the volume of water that would allow 16 grams of potassium chloride to dissolve. So what we have is the solubility of potassium chloride of 16 grams per 44.0440 millilitres. We can scale the denominator from 44.0440 mL to 100 mL by first dividing my 44.044 and then multiplying by 100. If we do the same to the numerator, we end up with a value of 36.3273 grams per 100 mL of solvent. So our final value for the solubility of potassium chloride at 15 degrees Celsius is 36.3 grams per 100 mL of water. Our final value is to three significant figures because we don’t actually use 15 mL in our calculation. We don’t actually use 15 degrees Celsius in our calculation. The only value that limits our significant figures is the value for the density of water, 0.999 grams per milliliter.