Use the values of Ksp in Appendix C to calculate the solubility of the following compounds (in g/L): (c) Cu3(PO4)2

Video transcript

Hi everyone. This problem reads if the value of the cell viability product constant for calcium phosphate is 1.0 times 10 to the negative, 25 what is its soluble itty and grams per leader. So we want to answer the question of so scalability and grams per leader. And in order to do this, what we're going to do is we're going to solve or calculate for moller solly ability. So that's our end goal is to calculate for X. Which is going to be our moller cell ability. And we're going to use an ice table in order to do this. Okay, But first let's go ahead and balance our reaction because it is not balanced. Okay, So what our balanced reaction is going to become is the following. So we have three calcium cat ions plus two phosphate ions. Okay, so this is our balanced reaction. So now we're going to construct an ice table. Remember that solids are ignored in the ice table. So we're going to write this but we're going to ignore our solid. So we have ice written here and we don't include solids in the ice table. So we're essentially ignoring this. And so we have no products. So the initial concentration is going to be zero for both, which means that our reaction is going to shift in the ford direction. Okay, so we're going to make product. So our change then becomes positive and we're looking at how many moles we have. We have three moles and our changes represented by X. And for our second product we have positive because we're increasing concentration. We have two moles of it and our changes represented by X. So that means then the equilibrium row for our ice table becomes three X. And two X. Okay, so now we can write out our equilibrium are scalability product constant expression, R K S. P expression. So R K S P is equal to and note that solids are ignored and the concentration is raised by the stoke E. O. Metric coefficient. So this is going to equal the concentration of products over the concentration of react ints. Okay, so let's just write out what this means. So from our balanced reaction it's going to equal our concentration of products which is the calcium cat ions raised to its Tokyo metric coefficient times the concentration of phosphate and ions raised to its Tokyo metric coefficient. And this is all gonna all be over one because we don't include solids in our equilibrium constant expression. Okay, so now we know what the value of or what we want to solve for. Or we are told the value of our solar cell ability of our yes, we're told the value of K. S. P. And in the problem we're told the value is one times 10 to the negative 25. So we have that. So let's go ahead and plug that in. So we have one times 10 to the negative 25 is equal to. Now we're going to plug in the equilibrium concentrations from the zero in our ice table. So this is going to be equal to the concentration of calcium or this one is the first one. So that's represented by three X. And we're raising raising it to the third power. And for the phosphate. And I on this is going to be represented by two X. And re raise this to the second power. Okay? So now we're going to calculate for X which is going to give us the moller solly ability. Okay, so let's go ahead and simplify here. So we're going to get one Times 10 to the -25 is equal to When we simplify the right side. This is equal to 108 Times X to the 5th. Okay, let me put this in parentheses so we can see that. All right. So now we're gonna divide both sides by 108. So let's go ahead and do that. Okay, so we get X. Rays to the 5th is equal to and we're also going yes. So we're going to get X to the fifth is equal to one times 10 to the negative 25th over 108. And we're gonna raise this to The 1/5. So we can get rid of the exponent here. Okay, And so when we do this then what we get as our final answer for X. Is X. Is equal to 3. times 10 to the negative six molar. So now that we know what X. Is, which is our moller solly ability. We can calculate the cell viability in grams per liter. Okay, because this X. Represents our moles of the calcium phosphate. So let's go ahead and finish this off. So we have 3.9 to 0 to six times 10 to the negative six moles of calcium phosphate per leader. Because that are moles is moles per liter. So now we want to go from moles per liter, two g per liter. And we're going to do that using the molar mass of calcium phosphate. And we'll need our periodic table to do this. So in one mole of calcium phosphate, the molar mass is 310.18 g. Okay, so our units for moles cancel. And now you can see we have grams per leader. So we just need to do the Calculation and we'll be done. Okay, so once we do that, we get an answer of 1.2 times 10 to the negative three g per leader. And this is going to be our final answer. That is it for this problem. I hope this was helpful