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Ch.17 - Additional Aspects of Aqueous Equilibria

Chapter 17, Problem 105

The solubility product constants of PbSO4 and SrSO4 are 6.3 * 10-7 and 3.2 * 10-7, respectively. What are the values of 3SO4 2 - 4, 3Pb2 + 4, and 3Sr2 + 4 in a solution at equilibrium with both substances?

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Hi everyone. This problem reads the cell viability product or K. S. P. Values for manganese to carbonate and copper to carbonate are 5.0 times 10 to the negative 10 and 2.3 times to the negative 10 respectively. In a solution that is at equilibrium with both compounds, calculate the concentration of the following ions. Okay, so we have three concentrations that we need to solve for. And so because we know we're at equilibrium with both compounds, we're going to need to write out our equilibrium. Okay, so let's go ahead and do that first. So for the first one we have manganese to carbonate and this is a solid. And at equilibrium it's going to have manganese ions as well as carbonate ions. Okay. And we were given the solid ability product value for this which is R K. S. P. And so we have our K. S. P. And this value is equal to the product of the molar concentrations of the ions. So we have our manganese ions and our carbonate ions. And the value for K. S. P. Was given in the problem, this value is 5.0 times 10 to the negative 10. Alright, so let's go ahead and do the same thing for our copper to carbonate. So this is going to be our solid. And at equilibrium the ions are copper ions. Let's write that a little clearer. So we have our copper ions and our carbonate ions. And we're also given the solid ability product or the K. S. P. For this which we said is the product of the molar concentrations of the ions and the value, the K. S. P. Value for this one is Given and it is 2.3 times to the negative 10. Okay, so our goal here is to calculate the concentrations for the carbonate ion, the magnesium ion and the copper ion. So we're going to need to create some hypotheticals here. So let's go ahead and do that. So we're going to say we're going to let X equal our concentration of manganese ions. We're going to let why equal our concentration of copper ions and we're going to let X plus Y equal our concentration of carbonate ions. Okay, so our goal here is to solve for X and Y. Okay, so our two equations are two K S P that we have, we're going to set them equal to what we have written on the side here. So for our first Caskey, Alright, A one here and this is two. Okay, so for our first one we have 5.0 Times 10 to the -10 is going to equal X times X plus y. We're substituting what we wrote out over here on the right, so if we look at our first one we have K. S P is equal to X times X plus Y. These are how we defined it. Okay, so for our second equation we have 2.3 times 10 to the negative is equal to Y times X plus y. Okay, so I'll write it here. So we have it as Y times X plus Y. All right, so I'll go ahead and erase this year. So now that we have these two set up and equal to each other we're going to divide. Okay this bye each other. So we're gonna have we're going to set it up as a proportion and set it and divide by each other. And when we do that you'll see that our X plus Y cancel. And we're just left. Now we can simplify. So the left side, once we simplify becomes 2.1739 and the right side equals X over Y. So let's define this in terms of X. So when we rearrange this to software X. We get X is equal to 2. 39. Y. Okay, this is very important. Okay, so this is our first important piece so we know what X is. Alright. And now we I have one more step which is to solve for y. Okay, so let's go ahead and solve for Y. And we're going to choose one of our equations are one of our key sp equations to be able to solve for Y. So let's go ahead and choose our second equation. Okay, so let me rewrite out our second equation here. So for a second equation. Alright, a two here we have 2.3 times 10 to the -10. It's equal to why times X plus y. Excuse me? I was just making sure that was clear. So our second equation is going to be Y times X plus Y. And we just saw for X. Okay, so we're going to plug in what we just solved for. Okay, so we're going to take this step by step 2.3 times 10 to the negative 10 is equal to Y times X. We know what X is. We know X is 2.1739, Y plus Y. Okay, so we have to wise here so we're going to simplify this. So this is this Y alone represents one Y. So now we're going to add those two. Okay, so we have 2.3 times 10 to the negative. 10 equals y times 3.1739. Why? Okay. And that's because we have a one in front of this Y. Right here. So 2.1739 plus one. Okay, and that's how we get that. 2nd. Why? So now we're going to simplify this even further because we have why times Y. Okay, so this becomes 2. times 10 to the negative 10 equals 3.1739. Y squared. Okay. And our goal here is to solve for Y. So in order for us to solve for Y, we first need to isolate it. So we'll divide both sides of our equation by 3.17 39. Okay, so once we do that and we isolate it. We get y squared Is equal to 7.246, 6 times 10 to the -11. So are y is still squared and we want to get rid of that. So we're going to take the square root of both sides. Okay, so that means our Y is going to equal the square root of 7.2466 times 10 to the negative 11. Alright, so we get why once we take the square root is going to equal 8. times 10 to the negative six moller. And remember we defined why as it is equal to our concentration of copper ions. Okay, so that means our copper concentration is equal to this number. So 8.5127 times 10 to the negative six. So this is our first answer we just solved for because up at the top we said we're going to let y equal our concentration of copper ions. Okay, so we know why which means we can now solve for X where X represented our concentration of manganese ions. Okay, so we're almost done here. So we said X Let's do a different color, let's do black. So we said X is equal to 2.1739 Y, correct. And now we know the value of Y. So we have 2. times are Y. is 8.5127 times 10 to the -6. Okay so that means our value is 1.9 times 10 to the negative five moller equals our concentration of manganese ions. Okay so let's just write that out here. Our concentration of manganese ions is equal to 1.9 times 10 to the negative five. Okay, that's our second answer. And lastly we said that our concentration of carbonate ions is equal to X plus Y. So let's go ahead and plug in those values. So we said X is 1. times 10 to the negative five plus Y. R. Value for Y is 8.5127 times 10 to the negative six. This gives us 2.7 times 10 to the - Moller. So that means our final concentration, Our final answer that we're solving for the concentration of carbonate ions is equal to 2.7 times 10 to the -5. Let's rewrite that times 10 to the negative five molar. Alright, so these are three concentrations were asked to solve for and were able to do it by using substitution and that is it for this problem. I hope this was helpful