The solubility of copper(I) chloride is 3.91 mg per 100.0 mL of solution. Calculate Ksp for CuCl.

Video transcript

Hello, everyone. Today, we have the following problem. Caesium nitrate has a solid ability of 0.259 g per middle leader, determine the value of K S P for caesium nitrate. So first thing that we want to do is you wanna draw out this dissolution reaction. So we have our cesium nitrate in the solid form and it's gonna react in a reversible reaction to form cesium plus ions as well as nitrate Aquarius ions. Then of course, we do want to make a note of the molar mass of caesium nitrate And that is going to be equal to 194. g per mole. And this can be taken by taking the molar masses of one cesium from the periodic table and adding that to the molar mass of one nitrogen from the periodic table and then adding that to three times the mass of oxygen on the periodic table and adding those together to get this molar mass here. And so we have the ability in the questions itself and we can go ahead and answer the question for the concentration. So to find the concentration, which is the first step in this problem. We're going to take our salt ability of . grams per middle leader. And then we're going to multiply that by the molar mass. And we can say that we have one mole And in one mole, we have 194.9 g. Then of course, we want to get rid of our military leaders and turn it into leaders. So we're gonna use the conversion factor that 10 to the third ml is equal to one leader. And of course, our units of Middle L and g will cancel that will be left with 1.3 moles per liter, which translates to 1.3 molar since our units of polarity is in units of moles per liter. And so next, we need to create an I C E table for this reaction. So we have our caesium nitrate in the solid form in our reversible reaction forming our cesium ions and our nitrate. Then we have our ICU table. We have our initial change in our equilibrium solids are not included in this table. So we're going to skip over that. Our initial concentrations for our products will of course be zero. However, since we are adding to them, they're going to be positive X and R E Rowe is just going to be our initial plus our change. So we're just gonna have plus X and plus X. So we can now write out what R K S P would be. So R K S P value is essentially going to be our products over our reactant. However, since caesium nitrate is a solid, it will not be included in this case, the expression. So we essentially just have our cesium times our nitrate. And so we note that we, we denoted that cesium was denoted by X and nitrate was noted by X. So therefore, X is going to equal the concentration of cesium, which is going to equal the concentration of nitrate. And we've already determined that that was going to be 1. malaria. So what we're gonna do is we're gonna take this K S P expression and we're going to apply it. And so we're going to have our 1.3 molar Times our 1.3 Molar. And this will give us a KSP expression of 1. as our final answer. And with that, we've answered the question overall, I hope this helped. And until next time.