Equilibrium in Aqueous Solution Solubility Equilibria the Solubility Constant Product [Page 1 of 2]

Equilibrium in Aqueous Solution Solubility Equilibria The Solubility Constant Product [Page 1 of 2]

We have quite a menagerie of equilibrium constants now. We have Kc, Kp, Ka, Kb, Kw. Now we’re going to add another one: Ksp. It’s the solubility product, and it governs the solubility of sparingly soluble salts. Let’s first talk about what a sparingly soluble salt is. If you take table salt, sodium chloride, you can start sprinkling it in, and it’ll dissolve. And you can actually dissolve a considerable amount: more than 5 moles of sodium chloride in a liter of water. What we’re imagining now is a different compound—say, chalk, calcium carbonate, which is a sparingly soluble salt. When you drop a piece of chalk into water, for the most part it, just sits there. That’s good for oysters, because their houses are made of calcium carbonate. Now, we’re not talking about a kinetic phenomenon. It’s not that it’s really slow to dissolve, but rather that the sodium chloride will be freely dissociable to an extent in the water, and the calcium carbonate will never dissolve to any great extent in the water. The reason is because the solubility product for calcium carbonate is relatively low.

Here’s a table that has some examples of solubility products. You just look these up in a table. You’ll note, first of all, that they’re temperature-sensitive, so in other words, they’re listed at 25°C, but just like any other equilibrium, they change with temperature. For the most part, they’re pretty small, so they’re much, much less than 1. You have different compounds having different solubility products. Now, there’s a distinction between the solubility product and the solubility, and that’s the concept that we’re going to be talking about in this lecture.

The solubility product is just an equilibrium constant. We have silver chloride, a sparingly soluble salt. When it dissolves, we get some silver ions in solution and some chloride ions in solution. And the equilibrium expression, just as always, is going to be the concentration of silver ions times the concentration of chloride ions. That’s our solubility product. For something like calcium phosphate, that’s a little more complicated because it consists of three calcium ions and two phosphate ions, we’re going to have that the equilibrium constant is going to be the calcium-ion concentration cubed (where this 3 translates to this 3), and the phosphate-ion concentration squared (where this 2 translates to this 2). Keep this in mind, the silver chloride equilibrium, because we’re going to be using it to solve a problem.

The problem we’re going to solve is: what is the solubility of silver chloride in water if the solubility product is 1.77 x 10–10? Again, solubility and solubility product are two different things. I’m going to define the solubility of silver chloride to be equal to s. If you think about it, if s is the solubility of silver chloride in moles per liter, then for every mole of silver chloride that dissolves, we’re going to get a mole of chloride ions and a mole of silver ions. So at equilibrium, the concentration of chloride ions will exactly equal the solubility, and the concentration of silver ions will exactly equal the solubility. This, of course, is going to be a small number. Since Ksp is just the product of the silver-ion concentration –10 times the chloride-ion concentration, we have 1.77 x 10 is equal to Ksp, which is equal to the product of these two things, which is just equal to s2. And it’s a simple matter to take the square root to find what s is. In this case, s is 1.33 x 10–5 molar. What that says it that if we had 1.33 x 10–5 moles of silver chloride, that would dissolve in one liter of solution, one liter of water, and give rise to a concentration of silver ions that was also 1.33 x 10–5 molar and a concentration of chloride ions that would be 1.33 x 10–5 molar.

This problem is a little more complicated if we consider calcium phosphate. You’ll recall that when calcium phosphate dissolves, you get 3 moles of calcium and 2 moles of phosphate, so the equilibrium expression is going to be a little more complicated. Go back and look and see what the solubility product expression is for calcium phosphate. When we let s equal the solubility of calcium phosphate, though, what you’ll note is that there is no concentration of ions that is exactly equal to s, because for every mole of calcium phosphate that we dissolve, we get 3 moles of calcium ions and 2 moles of phosphate ions. So at equilibrium, the concentration of calcium ions is going to be 3 times s, and the concentration of phosphate ions is going to be 2 times s. But we can plug these expressions in for the calcium ion and 3 2 5 the phosphate ion. We get (3s) (2s) , which is 108s equal to Ksp. Then it’s a relatively simple matter to divide by 108 and take the fifth root, and we get that the solubility is equal to 1.14 x 10–7 molar. Again, this doesn’t represent either the concentration of calcium ions or the concentration of phosphate ions in solution. What it represents is the number of moles per liter of calcium phosphate that will dissolve. And again, it’s not the same as the solubility product, because solubility and solubility product are two different things.

One caveat that I need to mention is that the solubility of these sparingly soluble salts that involve very basic anions— and phosphate is pretty basic, even though it’s a weak anion, it’s a pretty basic anion—these numbers are probably going to be a little bit off, maybe by as much as an order of magnitude. The reason (and we’ll talk about this more in a Copyright © Thinkwell Corp. All rights reserved. www.thinkwell.com

Equilibrium in Aqueous Solution Solubility Equilibria The Solubility Constant Product [Page 2 of 2] future lecture) is because phosphate ion is a fairly basic anion, so it reacts with water, and there’s an additional equilibrium that can affect the solubility of calcium phosphate. The reason why I chose this problem is because the algebra is about as complex as it’s going to get. If you can understand how to do the calcium phosphate problem, you should be able to do all the other problems, because they’re going to be easier than the calcium phosphate problem.

We can turn this problem on its head and look at it from the other way. Suppose we know the solubility of something— say, the solubility of calcium fluoride—and we experimentally determine it to be 2.1 x 10–4 molar. Then we can calculate what the solubility product is. Because we know that for every mole of calcium fluoride that dissolves, we’ll get a mole of calcium and 2 moles of fluoride, we can let the solubility be the concentration of calcium ions, and then the concentration of fluoride ions is going to be 2s. And the solubility product is still the calcium-ion concentration times the fluoride-ion concentration squared, which is s(2s)2, which is 4s3. So it’s just a matter of…since this is s, we take this number, we cube it, we multiply by 4, and we get 3.7 x 10–11. You can look up what the actual solubility product of this compound is, and you’d find that it’s listed as 3.45 x 10–11. So we’re within about 10% or so of the experimentally determined value, based on our experiment that we did in the lab.

Finally, it’s important to understand that you can compare solubilities sometimes. You have to make sure that you’re comparing apples and apples, and not apples and oranges. What am I talking about? We have silver chloride, silver bromide, and silver iodide, and we have three different Ksp’s, three different solubility products. Because the ratio of anions to cations is the same for all these, it’s going to be the case that the solubility is related to the solubility product in exactly the same way. In other words, for each one of these, the solubility is going to be given by the square root of the solubility product. So we can, at a glance, compare these three numbers, see that the solubility product of silver iodide is much less than the solubility product for these other two compounds, and by extension, we can say that silver iodide is the least soluble of these three salts.

But you have to understand that if we compare something that has a different relative formula—in other words, say, silver phosphate, which has three silver ions and one phosphate ion—its solubility product might be much smaller than the solubility product of all these other compounds, but you cannot, just looking at this, know that the solubility of silver phosphate is much less than the solubility of silver iodide. In fact, the solubility of silver phosphate is much 2 3 greater than the solubility of silver iodide, because Ksp for silver phosphate is not equal to s but is equal to (3s) s. You can go back and look at what we did for calcium phosphate, and by extension, you can figure out that this is the solubility expression for silver phosphate. If you solve this, what you find out is that the solubility of silver phosphate is actually as large as the solubility of silver chloride, so it’s actually about as soluble as silver chloride, even though its solubility product is much less than silver chloride and much closer to silver iodide. Again, the reason is because the relative number of anions and cations is different, so your solubility product expression is going to be different. So you can compare things where you have the same relative formulas, but you cannot compare things that have different relative formulas.

What is all this going to be useful for? Later on, what we’re going to learn is that these are really useful for allowing you in the lab to separate mixtures of solutions of ions based on their different solubilities.