Here we have lead to fluoride is a white solid and has diverse applications in pharmaceuticals, metallurgy and technology. If the concentration of lead to fluoride is 4.2 molar with a K_{sp} of 3.6×10-8, calculate the molar solvability of the solid at 25°C. All right, So when they say solubility, whether they say calculate solubility or calculate molar solubility, they're just asking you to find X. And when they're asking for the most ability of the ionic solid, that's just equal to X.

Here we're going to follow the steps in order to solve for this particular question. So step one, we set up an ice chart with solid as the only reactant because we were looking at how it breaks up into ions cross out the reactant side because remember in a nice chart we ignore solids and liquids. This is an ice chart, so this is initial change equilibrium using the initial role set products equal to 0. Now remember we lose reactants in order to make products, so using the change row place a + X for the products. We also have to take into account coefficients, so here this would be plus X. There's a 2 here, so this would be plus 2X.

Using the equilibrium row, we're going to set up the equilibrium constant expression with K_{sp} and solve for X. So now this is plus X and this is +2 X. Now here we're going to say that the variable X in the ice chart represents the molar solubility of my ionic solid. All right, so here we're going to say K_{sp} equals products over reactants. But again, the reactant is a solid so we ignore it. So it just equals PB^{2+} or plus 2 * F^{-}. Remember, the coefficient also pops up here too as a power.

All right. So now we're going to plug in values that we have. So we're going to say here K_{sp} is 3.6×10-8 at equilibrium led to his X fluoride is 2X, and that's still squared. So we're going to say this is X times 2^{2} is 4, X^{2} is X^{2}, so four X^{2} * X comes out to 4X^{3}, and that's still equal to my K_{sp}. So all we have to do now is solve for X. So divide both sides by 4 and we'll get X^{3} = 9.0×10-9. Take the cube root of both sides. When we do that, we get X = 2.08×10-3 and that'll be molarity. So this represents the molar solubility of my ionic solid.

So again, if any time they're asking for the molar solubility or solubility of your ionic compound is just equal to X. Now if they were asking for one of the ions though, it wouldn't just simply be X. What you would need to do is you need to look at the equilibrium row and then look to see what is that ion represent at equilibrium. Here led to his ex so it would still be this number but fluoride ion is 2X which would mean that you take this X answer and plug it into this 2X so it would be 2 * X and that would be the true molar solubility of this fluoride ion. So again be careful with when it comes to the ion. For the ions you have to check the equilibrium row and see what their equation is for the solid when you solve for X. That is your final answer.