Solubility Equilibrium - Concept

So in this segment let's go ahead and continue our discussion of equilibrium and talk about solubility equilibria. So let's define it first, so basically solubility equilibria is the equilibrium associated with dissolving solids in water to form aqueous solutions. So you probably have encountered this kind of a little bit in your journey talking abut just different types of reactions in general. So let's look at a specific example here, so barium sulfate here as a solid if you dissolve it in water, will break up into the ionic species of barium 2+ and sulfate 2- so initially when you put this solid in water, it's not going to dissolve right away so you still have quite a bit of barium sulfate solid left.

But over time, the concentration of the individual ions will start to increase so you'll get an increase in the concentration of barium 2+ and an increase in the concentration of sulfate 2-. So as you get a build up of these ionic species the likelihood starts to increase so they're going to interact with each other to go back and form the solid again. So at some point ultimately an equilibrium is reached. So again when a forward so the rate at which barium sulfate dissolves to form barium ions and sulfate ions is the same as the rate at which barium ions and sulfate ions form barium sulfate solid back.

Okay so at the point where no more solid is dissolving the solution is said to be what we call saturated. So no more solid can dissolve when the solution is saturated, so let's go a little bit deeper and look at this in terms of the equilibrium expression with which we're pretty familiar. So here for this one since we're talking about solubility the equilibrium expression is going to include this solubility product constant. So we're familiar with keq and kc and kp as equilibrium constants so for this one, for solubility it's the solubility product constant. But it's mostly formed the exact same way as the others. So we still have products over reactants. So here our products are barium 2+ ion and sulfate 2- ions and it would be over barium sulfate but we're also going to remember that for solids they're not involved in the equilibrium expression. So therefore, here it is omitted meaning ksp is only dependent on the concentration of barium ions and sulfate ions in solution.

Alright so let's kind of do an example with some numbers, so you could be asked to calculate the ksp value for barium sulfate and you could be given that it has a solubility of some value here 3.9 times 10 to the 5 moles per liter at 25 degrees C. Remember that your equilibrium constants are always defined at a specific temperature. So 3.9 times 10 to the minus 5 moles per liter or big M I broke out the units here so that I can make a point which is that. What this means for solubility is that 3.9 times 10 to the minus 5 moles of barium sulfate dissolve per 1 liter of solution. Okay, so therefore, we know that if this is the amount of barium sulfate that we have when it breaks up into each of it's ionic species the barium 2+ and the sulfate 2- that means that we have 3.9 times 10 to the minus 5 moles of barium moles per liter of barium 2+ ion and 3.9 times 10 to the minus 5 moles per liter of sulfate ion. So we can go ahead and plug that into our equation, our equilibrium expression so then we have ksp is equal to basically 3.9 times 10 to the minus 5 squared which will give us the ksp value for this particular scenario of 1.2 times 10 to the minus 9.

Again remember that your equilibrium constants are unit lists and that's solubility equiribria.