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). (b) Which cation precipitates first?

Hello everyone today we are being asked to determine the Cali on that will precipitate first. If potassium carbonate is used to precipitate one of the cal eons in a solution that contains 10.25 molar of manganese and 0.5 Mueller of sync. Were then asked to calculate the minimum concentration of this potassium carbonate that will start the precipitation of that carry on. So the first thing we wanna do is we want to draw our potassium carbonate being dis alluded In solutions. So we're going to say this is going to do this, is going to dissociate into two potassium plus ions as well as one carbonate to -1. And then we're also going to say that the concentration of our K two c 03 is going to equal the concentration of regular carbonate ions. And then we have to calculate that concentration of carbonate ions. So when we take our potassium carbonate and will react it with one of the ions, let's say we're gonna react it with meghan is first we're gonna say M two plus Mn two plus this is going to form manganese carbonate Solid as well as two potassium ions. We're then going to take this mega knees carbonate and we're going to dissociate to form our ice table. So we're going to say this solid is going to dissociate into mega knees two plus Aquarius as well as carbonate to minus Aquarius. And now we can set up our ice table here. So we have our rice table, we have our initial concentrations for the solid not mattering because it is a solid. But for mega needs we have 0.025. And for carbonate we have zero. We have a change in plus X because there's only one of each. And so our final is going to be 0.25 and X. And the reason why we don't say 0.25 plus X for magazines is because X here is negligible. So we don't it doesn't matter in the situation. The next thing we're going to do is we're gonna set up our K. S. P. Value. So R K S. P. Value for manganese Carbonate is going to be 2.24 times 10 to the -11. And when we set it up with our with our equation here, we're gonna see the concentration of manganese two plus it's going to be multiplied by the concentration of carbonate to minus. We plug in our values 2.24 times 10 to the -11 is going to equal our concentration of manganese which is .025. We're gonna multiply that by X, Dividing both sides by 0.025, we get that X. Is equal to 8.96 times 10 to the -10. In other words X. is equal to r carbonate concentration. And so we move on to our second portion where we have to react this potassium carbonate with zinc. And so we have potassium carbonate here in the form just as before, we're going to react it with zinc two plus Aquarius, we're going to form zinc carbonate, Aquarius or a solid. Excuse me and two potassium ions. We're going to take this solid that we formed this zinc carbonate and we're going to say that this is going to dissolve into one zinc two plus ion as well as a carbonate to minus ion. And then we're going to set up our ice table. So we're gonna have I. C. And E. Once again for solid, it doesn't matter. So we're going to just draw an X through their our initial concentration from that zinc was found in the question stem which is 0.50 moller for carbonated zero. Once again, our change is going to be plus X since we only have one of each of these ions. And so this is going to be 0.5 and X. And once again X is negligible here. R K S. P value for zinc carbonate is going to be equal to 1.46 times 10 to the negative 10. And so we set up our K sp table, we'll have concentrations of zinc two plus times our concentration of carbonate. We set up our equation furthermore, we have 1.46 times 10 to the negative 10 is equal to our concentration of zinc which is 0.5 times X X. is going to end up equaling 2.92 times 10 to the -9. And of course in a situation X is equal to carbonate. And so we have to find the minimum concentration, so we have to find the lowest number and so between our lowest numbers between reacting with manganese and zinc, we can confidently say that the manganese value is going to be the lowest concentration, so therefore we can say that manganese will precipitate or form a solid, so precipitate first. And so that is going to be how we reach our final answer. I hope this helped, and until next time.

Ch.17 - Additional Aspects of Aqueous Equilibria

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Brown 14th Edition

Ch.17 - Additional Aspects of Aqueous Equilibria

Problem 74b

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Chapter 17, Problem 74b

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). (b) Which cation precipitates first?

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Identify the possible precipitates that can form when Na2SO4 is added to the solution containing Ba2+ and Sr2+. The potential precipitates are BaSO4 and SrSO4.

Write the solubility product constant (Ksp) expressions for both BaSO4 and SrSO4. For BaSO4, Ksp = [Ba2+][SO42-]. For SrSO4, Ksp = [Sr2+][SO42-].

Look up the Ksp values for BaSO4 and SrSO4 from a reliable source. These values indicate the solubility of each compound in water.

Compare the Ksp values. The compound with the lower Ksp value is less soluble and will precipitate first when the concentration of SO42- is increased by adding Na2SO4.

Determine which cation forms the precipitate first based on the comparison of the Ksp values. The cation associated with the lower Ksp value 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 a numerical value that represents the equilibrium between a solid and its ions in a saturated solution. It is specific to each ionic compound and is used to predict whether a precipitate will form when two solutions are mixed. The lower the Ksp value, the less soluble the compound is, which means it will precipitate out of solution first when the ion concentrations exceed the Ksp.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form an insoluble salt, or precipitate. This process is driven by the formation of a solid that is less soluble than the ions in solution. In this scenario, the addition of Na2SO4 introduces sulfate ions, which can react with Ba2+ and Sr2+ ions to form barium sulfate (BaSO4) and strontium sulfate (SrSO4), respectively.

Comparative Solubility of Sulfates

The comparative solubility of sulfates is crucial for determining which cation will precipitate first. Barium sulfate (BaSO4) has a much lower Ksp than strontium sulfate (SrSO4), indicating that BaSO4 is less soluble and will precipitate out of solution before SrSO4 when sulfate ions are added. Understanding the relative solubility of these compounds allows us to predict the order of precipitation in the reaction.