A solution of Na₂SO₄ is added dropwise to a solution that is 0.010 M in Ba²⁺(aq) and 0.010 M in Sr²⁺(aq). (a) What concentration of SO₄^2- is necessary to begin precipitation? (Neglect volume changes. BaSO₄: Ksp = 1.1⨉10^-10; SrSO₄: Ksp = 3.2⨉10^-7.)

Video transcript

Hi everyone. This problem reads a solution contains both 0.15 moller of manganese ions and nickel ions. If potassium carbonate is added drop wise the solution. What is the mill arat e of carbonate ions when the second cat ion starts to precipitate. And were given the soluble itty product constants for both. Mag, unease to carbonate and nickel to carbonate. So let's go ahead and get started. So our goal here is to calculate the similarity of carbonate ions when the second cat ion starts to precipitate. So what we're going to need to do is figure out which one of these is going to precipitate first and we're going to have to do it one by one. Okay, so let's go ahead and start off with arm agonies to carbonate. Okay, we're going to write out the equilibrium for it. And that is we have our solid manganese to carbonate. Is going to yeah equilibrium. And we're going to have the ions manganese and carbonate. Alright. And we're given the soluble itty product constant K. S. P. And this is the product of the molar concentrations of the ions. Okay, so it's going to be the product of our manganese ion and our carbonate ion. And we were given this value and that value is 2.24 times 10 to the negative 11. And our goal here is to find out what is the concentration of carbonate ions here. Alright, so let's go ahead and start off by isolating this variable, this is what we're solving for. So that means we're going to need to divide both sides of our equation by our concentration of manganese ions. Okay, so we can isolate that. Okay, and when we do that we get our concentration of carbonate ions is equal to 2.24 times 10 to the negative 11 over our concentration of manganese ions. And we know what that value is because it was given in the problem. So let's go ahead and plug in. So we have 2.24 times 10 to the negative over 0.15. Okay, so that value for our concentration of carbonate ion is 1. times 10 to the negative nine moller. Okay, so that's our concentration of carbonate ions from our manganese to carbonate. Okay, so let's go ahead and do the same thing to figure out what our concentration of carbonate ion is from our nickel to carbonate. Okay so we have our nickel to carbonate, we're going to establish an equilibrium and it's going to have its ions nickel to nickel and carbonate. And we're going to do the same thing here. We want to calculate what is our concentration of carbonate ions. So let's write out our K. S. P. So R K S P is going to equal the product of the molar concentrations. Okay and we were given that value and the problem and that value is 1.42 times 10 to the -7. Okay, so now we want to solve for our concentration of carbonate. So go ahead and divide both sides of our equation by our concentration of nickel ions. Okay, Okay, so that gives us our concentration of carbonate ions is equal to 1.4, 2 times 10 to the negative seven, divided by our concentration of nickel ions. And we know that value because it was given. So we have 1. times 10 to the -7, divided by 0.015. And when we do this calculation we get 9.4667 times 10 to the negative six. Okay, and this is moller. Alright, so based off of these two values because our concentration of carbonate and our manganese to carbonate is smaller, it's going to precipitate first. Okay, Okay, so that means our nickel ion is the second ion to precipitate and it will precipitate when the carbonate ion is equal to 9.4467 times 10 to the negative six molar. This is what the question was asking us for. What is the mill arat e of the carbonate ion when the second cat ion starts to precipitate. Okay, so that is the answer to this problem and that is the end of this problem. I hope this was helpful