Precipitation Reactions - Using the Solubility Rules

Basic idea of precipitation reactions. NOTE: when working with precipitation reactions, the solubility rules for ionic compounds are used to determine if a precipitate forms or not.
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[@lucasthethird]

Ben--I have watched about two hours' worth of videos explaining precipitation reactions, aqueous solutions, net ionic equations, and so on. Yours is by far--and I mean by FAR!--the clearest and easiest to understand, with the most well-laid out and clarified examples. Thank you so much!

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Thanks Luke. Means the world to read such a kind comment!

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Thanks Ben! You're still helping students 9 years later! Very helpful!

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Thank you for your kind words, and thanks for watching! 😀

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Thanks so much! I'm having the hardest time in Chemistry, and I've been having to look up a lot of "How to" videos. Thanks for making things simple and easy to understand! :]

[@Finnbrack]

I had no idea how precipitation reactions worked before watching this video, pretty darn confident now, god bless you, sir.

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Its now 2018 and this video has still proven to be useful thanks

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Thank you so much! Your explanation is so clear.

[@TheCrazyStats]

Thank you. I finally understand precipitate much better

[@28princesst]

Thanku so much, you make it seem so easy, and yes our teacher is making us memorize that hold chart, along with limiting reactant problems, and finding the emperical formulas from the limiting reactant, and oxidation-reduction rules, not to mention all of the Gas Laws. Some body please pray for me and my classmates!!! But i thankyou for helping me with a clear understanding of precipitaion reactions. Sorry for the paragraph but i had to vent to someone lol

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[@chipnwafersticks]

Thanks for this. It helped me. I have a reporting in our chem. class tomorrow and I need to discuss this and double displacement reaction. If not for this video, I wouldn't understand precipitation reaction.

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Thank you thank you thank you!

[@ArtgirlRusher776]

This makes so much sense! Thank you.

[@xkarnx2355]

I wanted to ask from that last question that's Na2SO4 + KCl. When switching how come that so4 doesn't come first and having to determine if it is aqueous or not. Should it be So4K(aq)?

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[@hallcory1993]

If you look at the periodic table of the elements, whatever column that the element is in correlates if it is positive or negative.

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This was such a big help! Thank you!

[@ManuelMaciasHouTex]

why did you swap the NO3 and the Br and how do we know whether it is to be swapped or not?

[@ipaad]

Thanks for the help.

[@trayzzay85]

QUESTION!!!!! doesn't the last equation 9:55 have to balance yet? Since it isn't balanced will that affect the outcome? I don't see how it would. IDK

[@simphiweciliza4727]

Thanks man. Helped me BIG TIME!

[@molly1376]

Why do the pairs switch in the reactions? As in, why are the cations of the reactants paired with the anions they weren't with initially, for the products? Sorry if my question is unclear but this is something I am having trouble understanding.

[@elishasantos582]

Molly Koch Its a chemical reaction called Double Displacement.

[@YoshiHDify]

Is there an easy way to remember the rules. Because in an exam they won't be giving you a clue or any rule sheet. Or do we just have to dead set remember them?

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You're very welcome! Thank you for your kind words! 😀

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Thank a lot ser!

[@lovedog49507]

If the compound has a group 1 element is it automatically soluble ?

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[@BensChemVideos]

Water is a liquid, so we use (l). Soluble ionic compounds dissociate in water, so we use (aq) for aqueous. Insoluble ionic compounds do not dissociate in water, so we use (s) for solid. Even if an ionic compound is soluble, if it's not in water then it's usually a solid at room temperature, in which case (s) would be appropriate.

[@myst4216]

Very helpful! Thanks!

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This is excellent. thank you

[@ALananhR]

Would it be possible for you to do a video(if you have not already) on writing a balanced formula equation, complete ionic equation, and writing a net ionic equation? I am learning it in class, and I am totally and completely lost

[@parveenshaikh4136]

Why the concentrations of the two solution in a precipitation reactions are different?

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Lifesaver!

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[@abbassyed0]

Appreciate it. Very helpful.

[@sugreevsran766]

thanks ben

[@ryanramlogan6512]

In the last example no precipitate was formed but the cations and anions still switched, why is that considered no reaction?

[@elishasantos582]

Ryan Ramlogan I think its not right to say that they dont have reaction. They have reaction but not precipitation reaction because you do not produce a solid precipitate.

[@perfect204]

Awesome vid

[@shinsekai1]

thank you

[@renee7880]

Sorry, I am new to this. Where did the the 2 go from Na in the last example?

[@Icecube88]

+LZLZLZ ZEE it's because you want it to be in its simplest form. you can reduce Na2Cl2 to NaCl if you divide both by 2. it's in empirical form. I'm guessing you want the compounds in that form when you do this stuff. N2Cl2 is in molecular form.

[@Icecube88]

+Icecube88 yeah, make sure you balance the compounds when you put the elements together. if they can be reduced, then you reduce. Pb(C2H3O2)2 + Na2SO4 for example. You pair up Pb and SO4. They are balanced, so move on. You pair Na2 and (C2H3O2)2. Acetate (C2H3O2) has a charge of 1-, and there are two of them. Na has a charge of 1+, and there are two of them, so they are balanced. Since they can be simplified though, you simplify them. You can divide both by 2, and that will give you Na(C2H3O2). That's the compound that you will use. So the whole thing together is PbSO4 + Na(C2H3O2). Then you go from there (if they are soluble or not and so on. Well I want to add that when you do go on from there, you will find that it is a precipitation reaction.

[@Icecube88]

also, you can't reduce polyatomic ions. only if they are in parenthesis (indicating that there are more than one) can you reduce a polyatomic ion..if you get something like Na2SO4 paired up together, then you keep it like that. that's sodium sulfate. sulfate is SO4 with a 2- charge and sodium is Na with a 2+ charge. If you get Na4(SO4)2, then you can reduce it to Na2SO4. notice that the polyatomic ion hasn't changed, so you're good.

[@charlielewis5893]

what if both solutions produced are insoluble?

[@rixon3984]

Then both are precipitates

[@Sarah-ch5vw]

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[@pauldargan8995]

But the main question that none of these videos address is why the new compounds form - rather than reforming the original compounds? Is it an enthalpic or entropic reason?

[@nickshep]

Great vid :)

[@Dan-gc3ke]

If one of them is insoluble, then no reaction would occur right?

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THANK YOU!!!!!!!!

[@breadgenie1232]

8:33 Wait why is there a 2 next to the sodium

[@anthonyheppr]

Top man

[@RicardoKaka01]

Just a small mistake, for the final equation on 10:27, there should be 2 moles of KCl and 2 moles of NaCl so that it balances out the equation. You need to have 2 K, 2 Cl, and 2 Na on both sides for it to be balanced.

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[@mattv11900]

No, because Na is a 1+ charge and there's two of them, and Cl is ALWAYS diatomic (which means it exists in two) and also has a 1+ charge. So Na2Cl2 is just NaCl. :)

[@johnnyappleseed4020]

thank you!

[@dylanjohnson2395]

How come in Example 3, you don't need the equation to be balanced?

[@elishasantos582]

Dylan Johnson There's no need to balance the equation for that case because we are just trying to figure out if those reactants would produce a solid precipitate or not. We need to balance equation if we are solving stoichiometry problems..

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Many thanks, my parents included.