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You choose to investigate some of the solubility guidelines for two ions not listed in Table 4.1, the chromate ion $\left(\mathrm{CrO}_{4}{ }^{2-}\right)$ and the oxalate ion $\left(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\right)$. You are given $0.01 \mathrm{M}$ solutions (A, B, C, D) of four water-soluble salts: $$ \begin{array}{lll} \hline \text { Solution } & \text { Solute } & \text { Color of Solution } \\ \hline \mathrm{A} & \mathrm{Na}_{2} \mathrm{CrO}_{4} & \text { Yellow } \\ \mathrm{B} & \left(\mathrm{NH}_{4}\right)_{2} \mathrm{C}_{2} \mathrm{O}_{4} & \text { Colorless } \\ \mathrm{C} & \mathrm{AgNO}_{3} & \text { Colorless } \\ \mathrm{D} & \mathrm{CaCl}_{2} & \text { Colorless } \\ \hline \end{array} $$ When these solutions are mixed, the following observations are made: $$ \begin{array}{lll} \hline \text { Expt } & \text { Solutions } & \\ \text { Number } & \text { Mixed } & \text { Result } \\ \hline 1 & \mathrm{~A}+\mathrm{B} & \text { No precipitate, yellow solution } \\ 2 & \mathrm{~A}+\mathrm{C} & \text { Red precipitate forms } \\ 3 & \mathrm{~A}+\mathrm{D} & \text { No precipitate, yellow solution } \\ 4 & \mathrm{~B}+\mathrm{C} & \text { White precipitate forms } \\ 5 & \mathrm{~B}+\mathrm{D} & \text { White precipitate forms } \\ 6 & \mathrm{C}+\mathrm{D} & \text { White precipitate forms } \\ \hline \end{array} $$ (a) Write a net ionic equation for the reaction that occurs in each of the experiments. (b) Identify the precipitate formed, if any, in each of the experiments. (c) Based on these limited observations, which ion tends to form the more soluble salts, chromate or oxalate?

We're going to look at some double displacement reactions. Okay, so the first one is with sodium chrome eight, Miss smith ammonium C 204 which is known as oxalate, obsolete. Okay. So those who will react together in a double displacement reaction where they basically search cat ions and and ions to then form sodium oxalate. Okay, so as you can see here it has a two plus charge or two minus charge. Sorry? So the sodium, we'll need two of them. Okay then we get ammonium chrome eight. And as you can see, chromite is also die violent. So will be quite similar. Okay, so what we're at now is to decide which one is the precipitate. Okay, So if we look at it here, this reaction had no precipitate. So everything here is Equus and this was the only mixture that no precipitates. Now let's move on to the next one. It's again with the sodium crewmates and we'll be reactively reacting it with silver nitrate. Okay, what will form here again is double displacement switching of Catalans and and ions to get sodium nitrate with silver chrome eight. Okay. And what we see here is a red precipitate. So we know all nitrates are soluble. But so we would suggest that the silver crow mates. This is a solid. Okay. And silent. Put sub scripts in front of the other ones. That just means they're acquis. Everything else is Aquarius. Next up will be sodium chromium again and this time it will be with calcium chloride. Okay. And double displacement again to make some salt which is definitely soluble. This is the sodium chloride used to season food and then we have some calcium chromite. Okay, So then we know that this one is a solid here because we know that you know chlorides are soluble. Next up it will be B and C. So we have ammonia oxalate with silver nitrate. Okay, so we know the old form. Another nitrate which will definitely be soluble in this case. That's ammonium nitrate. Okay then the silver crow mates, as we've seen. Even in um example too is not that's soluble so it's just a solid. Again, this is a silver oxalate. Next up will be be in D. So we'll take this ammonium oxalate with calcium chloride. And what we get again, we'll be ammonium chloride, which we know will be valuable cards are soluble. Then we have calcium oxalate which must then be the solid. Okay then the last one is C. And D. So silver nitrate with calcium. All right. What will form is a G. Is a silver chloride. It's a calcium nitrate. So, that's when it's If the trick is here Well, we know nitrates are actually more soluble than quad. So required. Here is the solid. All right, So we just used the rules for the comment that we didn't know would be would be um soluble, which would be the nitrates and most of the chlorides, all the nitrates and most of the clients. So if we had to choose between a cooler and nitrate acquired would be be solved and the solid one.

On this problem. For the first one there's no precipitate. So there's no net ionic because all the items are going to cancel. So for the next one There is a precipitate. Nitrous are always soluble. So the precipitate must be the silver chrome eight. So basically to write the net ionic, you're just going to write two moles of silver because there's a sub script of two on the silver added to one mole of chrome eight because there's no self grip on the chrome eight, we get silver chrome eight solid. So that's basically what the net ionic shows is just showing the ions turning into the precipitate. So for the next one and A C. L. Is table salt we know it's soluble. So therefore this must be the precipitate. So we're going to write the non ionic for that For the next one, the nitrate is always soluble. So therefore this has to be the precipitate. And we'll have to write an ionic for that one and just be sure you're getting the correct charge too on this. So this would be the charges on these ions and these will just be something you would have to memorize. So be sure that you know the charges on the poly atomic ions and the common ions. So for this one the ammonium chloride is soluble because that's also assault. So this one has to precipitate. So we're going to just write none on it for that For the next one the nitrate is soluble. So we know this is a precipitate and we're going to write an ionic for this one we can get rid of the twos in front of everything because we're just dividing everything by two, and we'll just get that.

In this problem I can write DDX in it. To the G plus four. HN 03 Will react to form three G. A notary plus and new Plus two H 20. This is B. This is C and AGN 03 plus. N A C L will react to form GCL plus any annual teddy and a D C L plus to Energy four oh Edge will react to form Eddie an STD hold to see L plus two H two. And To a GN- 03 plus. Any too Ash two or 3 Will react to form 82 As two or 3 plus two. Any N O T. And largely and largely. Just look at it carefully edgy to add to or three. Will you? A G two S plus S +03 and this is black in color. So according to the option A option B. It correct? I hope you understand the solution.

To determine if a precipitate forms. The first thing that you need to recognize is three cat ions are going to switch places. When the cat ion switch places, you will get potential products. If one of those products is insoluble, then it forms a precipitate, and we have a precipitation reaction. If both of those products are soluble, then no precipitate forms and no reaction occurs. So you need to review the Sala Bility rules in Table 7.1 of this chapter. If we have sodium carbonate and manganese to chloride and the cat ions, which places will get sodium chloride and manganese to carbonate? According to rule number six carbonates are going to be insoluble, such as this manganese to carbonate for the next one. If we have potassium sulfate with calcium acetate, we'll get calcium sulfate and potassium acetate balance. It will put it to their no precipitate forms because, according to rule, number four, sulfates are going to be soluble, except with a few exceptions. And according to rule number two potassium salts, they're going to be soluble, so no precipitate forms here for the next one. We have hydrochloric acid reacting with Mercury one acetate producing acetic acid and mercury one chloride. According to rule number three Mercury one chloride is an exception to all chlorides being soluble. So this is our precipitate the next one. We have sodium nitrate reacting with lithium sulfate. Well, balance it. Our sodium sulfate is going to be soluble. Our lithium nitrate is going to be soluble, so no precipitate forms. Sodium sulfate is soluble for a couple of rules, one being rule number two sodium salts are soluble and the lithium nitrate is going to be soluble because, according to rule number one, all nitrates are soluble than for Part E. We've got potassium hydroxide reacting with nickel to chloride, producing sodium chloride and nickel to hydroxide. Nickel to hydroxide is our precipitate, according to rule number five hydroxide Czar Insoluble. Then we have sulfuric acid. Reacting with barium chloride. Producing hydrochloric acid and barium sulfate and barium sulfate is our precipitate because rule number four says Although sulfates are soluble, barium sulfate is an exception, and it is insoluble


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