A solution contains one or more of the following ions: Ag⁺ , Ca²⁺ , and Cu²⁺ . When you add sodium chloride to the solution, no precipitate forms. When you add sodium sulfate to the solution, a white precipitate forms. You filter off the precipitate and add sodium carbonate to the remaining solution, producing another precipitate. Write net ionic equations for the formation of each of the precipitates observed.

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Hey everyone. And welcome back to another video. A solution contains a one or more of the following ions, silver cion, calcium cion and copper two cion. When you add sodium chloride to the solution, no precipitate form. When he adds sodium sulfate to the solution, a wide precipitate forms, you filter off the precipitate and add sodium carbonate to the remaining solution producing another precipitate. Write net ionic equations for the formation of each of the precipitates observed. Now we have four answer choices. They have variations of those net ionic equations. And we can immediately say that we have two out of the three ions present in the solution because we are forming two precipitate. Now let's start analyzing the problem. What about silver? Well, first of all, we add sodium chloride to the solution, right? And sodium is a non spectator. So we can consider chloride anion and we need to think about the solubility rules. The majority of chlorides are soluble except from silver C ions mercury. And also we can include lead two CS. So now in this list, we have silver, right? But no precipitate forms. Meaning we can conclude that there is no silver C ions present in the solution. So we can exclude options A and B. They have the correct nat equation, but we don't have any silver present. We can exclude these two options. Now, let's continue. Then we add sodium sulfate. Once again, sodium is a non spectator. Let's think about sulfates and the majority of them are soluble except from calcium. According to the solubility rules stunned him. Barium silver cations and led to cion. From this list, we actually have calcium, right. We are we don't need to worry about the silver cion because we already concluded that there is no silver. So from this list, we can conclude that there is calcium carry on in the solution. And finally, we are adding sodium carbonate. So carbonate is the anion of interest because sodium is a non spectator. And what do we know about carbonates? The majority of them are insoluble except from group one a elements. So let's say if we have sodium carbonated, soluble and the ammonium cion, so ammonium carbonate would be soluble as well. Meaning if if the majority of them are insoluble, we are left with copper carbonate which is a precipitate. So now what we can conclude here is that number three, there is copper T plus in the solution. And now we can write the net ionic equation starting with calcium. We need to include its aqueous state. Now, calcium cion will react with the salted anion because they have equivalent, but opposite charges, calcium gets a subscript of one. And there's one unit of sulfate, we're including a solid state. And finally, we have copper two plus aqueous, it reacts with carbonate. Once again, we have equivalent but opposite charges. So we add one unit of each. So there are no subscripts there implicit because we have a one for each. And now we are including copper, carbon, copper, two carbonate solid state. We have our two net ionic equations. We simply want to determine which one is the correct option C or D. So looking at the answer choices, we can conclude that option C is the correct option because it has the two net ionic equations of interest. Thank you for watching.

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Key Concepts

Solubility Rules

Precipitation Reactions

Net Ionic Equations