# Video: EC17-18-S2-Q32

EC17-18-S2-Q32

03:39

### Video Transcript

Calculate Ksp of calcium fluoride, CaF₂, given that its solubility is two times 10 to the minus four molar.

Let’s start by thinking about what Ksp means. The subscript sp stands for solubility product. And the K means we’re looking for an equilibrium constant. So let’s start by writing out a reaction equation. We know that we’re starting with calcium fluoride, CaF₂. We also know that to have an equilibrium constant, our reaction must be in equilibrium with something. When we dissolve calcium fluoride in water, we get calcium two plus cations and two fluoride anions. This two will become important later on.

It’s useful to insert state symbols into our equation as well. So the calcium fluoride, before it dissolves, is a solid. And the two products are both aqueous. Now, let’s remind ourselves how we calculate an equilibrium constant. For a general reaction, we calculate the equilibrium constant by using the concentrations of each product raised to the power of this stoichiometry divided by the same for the reactants. So let’s see if we can do this for our reaction. Our products are calcium cations and fluoride anions. One calcium fluoride molecule produces one calcium cation. So we don’t need to include a power for this concentration.

However, one calcium fluoride molecule will produce two fluoride anions. So we raise the concentration of fluoride to the power of two. The question is, do we need to divide this by the concentration of the calcium fluoride? The answer is no because the calcium fluoride is a solid. And solids don’t really have a concentration. We would consider this a constant, and we don’t need to include it in our equation. So now we have our equation for Ksp. And we just need to insert some values. We’re given the solubility of calcium fluoride in the question, two times 10 to the minus four molar. There’s a one-to-one ratio of calcium fluoride to calcium cations. So we can say that the calcium cation concentration is two times 10 to the minus four molar.

However, when we look at the fluoride anions, remember that one molecule of calcium fluoride makes two anions of fluoride. So we need to multiply two times 10 to the minus four by two. So let’s put these into our equation. Now we can simplify the fluoride concentration expression. From here, we can then work out the final answer. Of course, if you want to write the expression on the right-hand side in standard form correctly, the expression would be 1.6 times 10 to the minus seven, which gives us a final answer of 3.2 times 10 to the minus 11.