A solution of Na 2 SO 4 is added dropwise to a solution that is 0.010 M in Ba 2+ (aq) and 0.010 M in Sr 2+ (aq). (a) What concentration of SO 4 2- is necessary to begin precipitation? (Neglect volume changes. BaSO 4 : Ksp = 1.1⨉10 -10 ; SrSO 4 : Ksp = 3.2⨉10 -7 .)

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

Ch.17 - Additional Aspects of Aqueous Equilibria

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Brown 14th Edition

Ch.17 - Additional Aspects of Aqueous Equilibria

Problem 74a

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Chapter 17, Problem 74a

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.)

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Identify the relevant chemical reactions for the precipitation of BaSO_4 and SrSO_4: Ba^{2+} + SO_4^{2-} \rightarrow BaSO_4(s) and Sr^{2+} + SO_4^{2-} \rightarrow SrSO_4(s).

Use the solubility product constant (K_{sp}) expressions for each salt: K_{sp, BaSO_4} = [Ba^{2+}][SO_4^{2-}] and K_{sp, SrSO_4} = [Sr^{2+}][SO_4^{2-}].

Substitute the given concentrations of Ba^{2+} and Sr^{2+} into the K_{sp} expressions: 1.1 \times 10^{-10} = 0.010 \times [SO_4^{2-}] for BaSO_4 and 3.2 \times 10^{-7} = 0.010 \times [SO_4^{2-}] for SrSO_4.

Solve each equation for [SO_4^{2-}] to find the concentration required to begin precipitation for each salt.

Compare the [SO_4^{2-}] values obtained for BaSO_4 and SrSO_4 to determine which salt will precipitate first.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is an equilibrium constant that applies to the solubility of ionic compounds. It represents the maximum concentration of ions in a saturated solution at a given temperature. For a salt like BaSO4, Ksp is calculated as the product of the molar concentrations of its constituent ions, each raised to the power of their coefficients in the balanced equation. Understanding Ksp is crucial for predicting when a precipitate will form in a solution.

Precipitation Reaction

A precipitation reaction occurs when two soluble salts react in solution to form an insoluble compound, known as a precipitate. This process is driven by the formation of a solid that cannot remain dissolved in the solution. In this context, the addition of sulfate ions (SO4^2-) to a solution containing Ba^2+ and Sr^2+ ions can lead to the formation of BaSO4 or SrSO4 precipitates, depending on the concentration of sulfate ions and the respective Ksp values.

Ion Product (Q)

The ion product (Q) is a measure of the current concentration of ions in a solution compared to the Ksp of a potential precipitate. It is calculated similarly to Ksp but uses the actual concentrations of the ions present at any given moment. If Q exceeds Ksp, precipitation occurs; if Q is less than Ksp, the solution remains unsaturated. In this problem, calculating Q for BaSO4 and SrSO4 as sulfate ions are added will determine the concentration of SO4^2- needed to initiate precipitation.