The accompanying photo shows the reaction between a solution of Cd(NO 3 ) 2 and one of Na 2 S. (b) What ions remain in solution?

Hey everyone in this example, we need to complete and balance the given molecular equation and provide a balanced net ionic equation including faces. So our first step is to write out the ions present. What we have is the PB two plus cat eye on, we have the C two, H 302 minus an eye on so that acetate an ion. And we have specifically two moles of this acetate, an ion based on the subscript given in the compound name. Next for ammonium bromide, we have the NH four plus catalon and the B R minus an ion. And so now in order to form our product, we're going to pair up the opposite charges. So we pair up the acetate and the ammonium and then we would pair up the lead two plus and the B R minus an ion. And what we would form as our first product, which is going to be lead bromide, So PB BR two. And based on our sustainability rules for bromides, this is going to be an insoluble solid precipitate because it's bonded to lead two plus, which is one of the exceptions for Celje bility for bromides. Our second product is going to be the formation of ammonium acetate, so NH four C two H 302. And this is going to have an AQ label because all acetate are soluble according to our sustainability rules. Our next step is to make sure this is balanced and in order to balance this out, we would place a coefficient of two in front of our ammonium acetate as well as a coefficient of two in front of our ammonium bromide. And so now everything is completely balanced and this is for our molecular equation. So our next step before we can get to the net ionic equation is to write out the complete ionic equation which includes spectator ions. And so what we would have is all of the ions present. So we would write out PB two plus on our reactant side, which gets the label because recall that all ions are soluble plus two moles of our acetate an ion. So two molds of the c two H 302 minus an ion added to two moles of our ammonium cat ion as well as two moles of our b r minus and ion. And sorry, these both get the label. So this completes our ions on our reaction side. And now for our product side we would keep our solid compound as is, we can't break that up into ions because it's not soluble. And so this would remain as lead bromide, solid. And next we would have our two moles of our NH four plus cat ion And then two moles of our acetate an ion. And so now we want to recall that when we have the complete ionic equation, we want to identify everything that is matching as far as the ions on both sides of our equation. So recognize that we have two moles of our acetate, an ion on the product side as well as our reactive side. So we're going to go ahead and cancel these out because these are spectator ions. And next we also have two moles of ammonium cat eye on on the product side as well as the reactive side. So we would cancel these out as well. So everything else left behind is what we would have for our net ionic equation. And so carrying everything down, we should have PB two plus with the accu label still plus two moles of R B r minus an ion producing our solid product which is going to be the lead bromide or the lead to bromide solid precipitate. And so this will complete this example as our balanced net ionic equation including our phases. So I hope that everything we went through is clear. If you have any questions, please leave them down below. So everything boxed in yellow first represents our balanced molecular equation as well as our balanced net ionic equation for our final answers. If you have any questions, I'll just leave them down below and I will see everyone in the next practice video

Ch.4 - Reactions in Aqueous Solution

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Brown 14th Edition

Ch.4 - Reactions in Aqueous Solution

Problem 92b

Chapter 4, Problem 92b

The accompanying photo shows the reaction between a solution of Cd(NO₃)₂ and one of Na₂S. (b) What ions remain in solution?

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Identify the ions present in the reactants: Cd(NO3)2 dissociates into Cd^2+ and NO3^- ions, while Na2S dissociates into Na+ and S^2- ions.

Write the possible products of the reaction by considering the exchange of ions. The potential products are CdS and NaNO3.

Determine the solubility of the potential products. CdS is insoluble in water, so it will precipitate, while NaNO3 is soluble and remains in solution.

Identify the ions that form the soluble product. Since NaNO3 remains in solution, the Na+ and NO3^- ions will be present in the solution.

Conclude which ions remain in solution after the reaction. The ions that remain in solution are Na+ and NO3^-.

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Key Concepts

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

Ionic Compounds and Dissociation

Ionic compounds, such as Cd(NO3)2 and Na2S, dissociate into their constituent ions when dissolved in water. This process involves the separation of the positive and negative ions, allowing them to move freely in the solution. Understanding this dissociation is crucial for predicting which ions remain in solution after a reaction.

Precipitation Reactions

A precipitation reaction occurs when two soluble ionic compounds react to form an insoluble compound, or precipitate. In the case of Cd(NO3)2 and Na2S, the reaction produces cadmium sulfide (CdS), which is insoluble in water. Recognizing the formation of a precipitate helps identify which ions will no longer be present in the solution.

Spectator Ions

Spectator ions are ions that do not participate in the actual chemical reaction and remain unchanged in the solution. In the reaction between Cd(NO3)2 and Na2S, the sodium ions (Na+) and nitrate ions (NO3-) are spectator ions, as they do not form a precipitate and remain in the solution after the reaction occurs.

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Textbook Question

A 0.5895-g sample of impure magnesium hydroxide is dissolved in 100.0 mL of 0.2050 M HCl solution. The excess acid then needs 19.85 mL of 0.1020 M NaOH for neutralization. Calculate the percentage by mass of magnesium hydroxide in the sample, assuming that it is the only substance reacting with the HCl solution.

A 1.248-g sample of limestone rock is pulverized and then treated with 30.00 mL of 1.035 M HCl solution. The excess acid then requires 11.56 mL of 1.010 M NaOH for neutralization. Calculate the percentage by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the HCl solution.

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Textbook Question

Uranium hexafluoride, UF₆, is processed to produce fuel for nuclear reactors and nuclear weapons. UF₆ can be produced in a two-step reaction. Solid uranium (IV) oxide, UO₂, is first made to react with hydrofluoric acid (HF) solution to form solid UF₄ with water as a by-product. UF₄ further reacts with fluorine gas to form UF6. (a) Write the balanced molecular equations for the conversion of UO₂ into UF₄ and the conversion of UF₄ to UF₆. (b) Which step is an acid-base reaction?