Use the following solubility data to calculate a value of Ksp for each compound. (a) SrF2: 1.03 x 10^-3 M

Video transcript

Hello everyone today. We have the following problem. The salt ability of rubidium chloride is 5.79 moller calculate the value of K. S. P. Four rubidium chloride. So the first thing we need to do is we need to create what's known as an ice chart for the reaction. So the reaction is going to be rubidium chloride as a solid dissociating into rubidium plus acquis ions as well as chloride ions. We're gonna draw our I. C. E. The initial concentration for our solid is not going to be present. We're not gonna have anything for the solid actually because solids are not included in this chart here. However, the initial concentrations for our products are going to be zero and the change is going to be X. For each of these products positive X. Because we are gaining them. And then so our equilibrium is going to be the the addition of our first two roles. So now that we have our two variables we can then calculate for R K S p A R K S p is going to be products over reactant. However, the reaction is a solid and solid are not included in the K. S. P. Expression. So it is just going to be our rubidium and our chloride. So it's said that the liability everybody in chloride was 5.79. So what does this mean? This essentially means that our X value which represents rubidium is also going to represent The polarity of our chloride and that is going to be both equal to 5.79. So essentially this is just going to be X times X. When we do this, we get KSPA 5.79 times 5. to give us a final KSP of 33.5. And with that, we have answered the question overall, I hope this helped, and until next time.