Greetings. In this lecture, I'll give some examples of solution calculation and also show how solutions can be made less concentrated through a process of dilution. Let's begin by reviewing some of the different ways that solution concentration can be expressed. Don't forget that concentration tells us how much solute is dissolved in the solvent. There's all kinds of different ways people can express the amount of solute that is dissolved. And the choice that it is made for how to express this is often a choice of convenience. For example, if someone made the solution by measuring the volume of a liquid solute and also the volume of the solvent, then it might be very convenient for them to express the concentration as percent by volume because they know the volumes that they mixed. If the concentration is extremely dilute or very low, it might be more convenient to express the concentration as parts per million or parts per billion. You often see this type of concentration used when people are talking about pollution. We've already talked about using molarity in our chemical calculations, because that's very convenient when we are doing stoichiometry. But people also sometimes will express concentration as percent by mass. Or mole fraction of solute in the solvent. Or another unit that engineers like to use is molality. The two that we'll be focusing on in this lecture are percent by mass and molarity, shown here in red. But I will also quickly run through a couple of the other ways that solution concentration can be expressed. Mass percent, which is sometimes shown as %w/w, w standing for weight, I know some you don't like that I use weight and mass interchangeably, but scientists do it all the time. Is defined as the mass of the solute divided by the mass of the solution, and please don't forget that the solution includes both the solute and the solvent. So in the denominator here, I have the solution not just the solvent, times 100, because that gives us 100%. Other types of solution concentration expressions that are similar to percent include part per million, which is the same calculation except that we use times 10 to the 6th instead of times 100. And part per billion, which follows along the same lines. Again it's mass of solute divided by mass of solution, but this time it's times 10 to the 9th. So parts per billion is useful for the most dilute types of solution. So these units are relevant to common lab reagents and analysis, particularly in environmental sampling. Here's an example for us to try. We're going to calculate mass percent of sugar in this glass of iced tea. We're told that this 0.4 liter glass of very sweet tea, because we're in the South, contains three heaping teaspoons of sugar. Assume that the density of the sweetened ice tea solution is 1.0 grams per mL. And that a heaping teaspoon of sugar, which might not be a unit of measure you know, has a mass of 8 grams. Hopefully, you are comfortable with 8 grams. So, we need to calculate the mass percent of sugar in the iced tea. What do we need to do that? Well first, we need the definition of mass percent, which is shown here. So for this particular solution, the mass percent of the sugar is going to be the mass of the sugar divided by the mass of the solution, times 100%. So that's what we need, is that calculation. We also need to use the density. So we have the mass of the sugar total. We know there are three teaspoons and they each had a mass of 8 grams. So there's 24 grams total of sugar. And then we know that the mass of the solution is 400 grams. If we really want it to be technical, we would show that this only has two significant figures. So it's 4.0 times 10 to the 2nd grams. This remember was the density of the solution, so this number includes the mass of the sugar. The result then, is the percent mass of sugar is 24 grams of sugar divided by 400 grams of solution times 100%, which gives us an answer of 6% sugar by mass. Probably we should express that to two significant figures, because if I look up here, I see that I have two significant figures for all of my numbers. So let's say that it's really equal to 6.0% sugar by mass. [SOUND] Let's proceed by talking about how to prepare a very precise solution of a certain molar concentration. What would we need to do if we were in the laboratory and we needed to prepare a solution that was exactly 1.00 molar concentration of sodium chloride and water? The first thing we would do is find a volumetric flask. In this case, I would like to make one liter of the solution. That makes the math pretty simple. So I'm finding a one liter volumetric flask. I'm then weighing out 1 mole of my solid, in this case, sodium chloride has a molar mass of 58.45 grams per mole. So I need 58.45 grams of sodium chloride in my volumetric flask. In the next step I'm going to add water to the volumetric flask, up to the mark that shows one liter, which is right up here. [SOUND] Then I'll swirl the flask to mix the solute and the solvent, and I now have a very carefully prepared 1 molar solution of sodium chloride in the water. What if we're not making an entire liter of solution? And what if we have a different solid? For example, how many grams of potassium phosphate, which has a molar mass of 212.3 grams per mole, are required to prepare 800 millilitres of a 0.500 molar solution of potassium phosphate in water? How would we do that calculation? Well we would start with what we're given, wich is 800 mLs of solution. The next thing we would realize, is that we have 0.5 moles per liter, which is the same as 0.5 millimoles per millilitre. We can multiply that times the fact that 1 mole contains 1000 milimoles, and finally multiply the entire thing by the molar mass of the potassium phosphate, 212.3 grams per mole. All of my units here should cancel out. So my millilitres cancel. My milimoles cancel. My moles cancel, and that leaves me with units of grams, which is exactly what I was asked for in the beginning of the question. So the answer here is 84.9 grams of potassium phosphate. Do you remember that potassium phosphate is not only soluble, but that it ionizes in solution? What is the chemical reaction for the dissolution of potassium phosphate in water? Please go ahead and try to answer that now. Thank you for submitting your answer. Of course when potassium phosphate dissolves, it produces 3 moles of potassium and 1 mole of phosphate for every mole of potassium phosphate solid that dissolves. So if I started with 0.500 molar potassium phosphate solution, I could calculate that the solution concentrate of potassium cation is 1.50 molar. And the solution concentration of the phosphate anion is 0.5 molar. That calculation gave us the answer to this question. What is the molar concentration of potassium? And what is the molar concentration of phosphate ion in the solution? The total ion concentration is the sum of those concentrations. And that gives us what's called the colligative molarity, which is important for determining some types of solution properties. In this case it was 1.5 molar plus 0.5 molar, which gives us a total ion concentration of 2.00 molar. And that was just the fact that we made four ions when we dissolved the potassium phosphate. If we started with 0.5 moles of potassium phosphate and dissolved that in one liter, that would give us 4 times 0.5 molar, for our total colligative molarity. The Van't Hoff factor here is 4. It's just 3 plus 1.