Ksp - Molar Solubility, Ice Tables, & Common Ion Effect

This chemistry video tutorial provides a basic introduction into Ksp - the solublity product constant. It explains how to calculate molar solubility from Ksp in mol/L and g/L and vice versa. It also explains how to solve common ion effect problems.
Full 1 Hour 26 Minute Video:
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Direct Link to The Full Video:
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PDF Worksheet - 18 Questions:
bit.ly/3irio7b
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Chemistry 2 Final Exam Review:
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Chapter 15 - Video Lessons:
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Full 1 Hour 26 Minute Video:
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General Chemistry 2 Final Exam Review:
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[@TheOrganicChemistryTutor]

Direct Link to The Full Video: bit.ly/3GANYHU Chapter 15 - Video Lessons: www.video-tutor.net/aqueous-equilibria.html

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[@tulliacairns9327]

Starting with molar solubility, find the Ksp 0:23 - is for 1:1:1 2:33 - is for 1:1:2 6:33 - is for 1:3:2 12:12 - is for 1:3:2 (different) Starting with a Ksp, find molar solubility 15:36 - is for 1:1:1 18:24 - is for 1:1:2 21:34 - is for 1:3:1 + find the concentration 28:31- is for molar solubility, concentration, then solubility

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Professor Organic Chemistry Tutor, thank you for a clear explanation into Ksp- the solubility product constant in AP/General Chemistry. Once again, the exercises/practice problems are excellent, however there are errors in problem 2 and other problems in the video. Professor Organic Chemistry Tutor, please find and correct these errors in the video.

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There are a couple of these where the Ksp is in g/L so you have to convert to moles first for the equation to work

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[@leahmichieli2268]

🤔 Are you missing the x 10^-3 on problem #2 @ 4:24 ? Or am I missing something?

[@zoe-hc6nf]

yeah he is

[@Sunflower28384]

He actually included it in the end because 6.875 x 2 is 13.75 but made it 0.01375 because it’s x 10^-3.

[@KhurshidisHere]

6:19 Why is 6.875 and not 6.875*10^-3 when finding the molar solubility for OH-? Why could not just do 2(6*10^-3)? Then, why raise it to the power 2 when you already multiplied it by 2?

[@OmejeJonas]

Exactly my confusion

[@kimberlyknox7283]

wish there was one where the reactant was aqueous because those ice tables get me confused sometimes

[@trueKENTUCKY]

Bruh im in a 500 level class and having to retrain this thanks

[@brandij2396]

I'm confused. Some equations he multiplied the coefficient, then other times he made them exponents. Like with examples 3 and 4. How do you know when to do which?

[@WindMaid]

you multiply by the coeffecient when you are trying to isolate the concentration of one ion in a compound. the x values calculated by ksp = [ion][ion] is just for ONE each of that compound.

[@linsdesk7240]

35:29 timestamp

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sm spam in the comments of this holy content also so how do you know if it's one object in a compound instead of two. example. AlPO4 is aluminum and phosphate two objects instead of doing (x) (x) (x)^4 it's just (x) (x) How do you know?? i have like 6 missing assignments in chemistry I've been scraping by with a B- all year I work 60 hr weeks and istg without this amazing guy I would be failing.

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[@dragontatoes]

2:30 where did this number come from? What are we doing with the molar concentration to get 1.4x10^-7?

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Thank you

[@garfield215]

why is PO4 in 12:44 aqueous? i thought it's only soluble if it is mixed with the group I elements?

[@Troy_1203]

YES GROUP 1A AND NH4. HOWEVER, IN THIS CASE PO+4 IS AN ION AND ALL IONS ARE IN (aq)

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How would you solve a question like this? You are studying a 3:1 salt. What is the Ksp if the molar solubility is (2.51x10^-4) M?

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In example 4, why did you not multiply the molar concentration of Ba2+ by 3, and the concentration of PO43- by 2, before raising them to their respective powers?

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the problem text gave the Ba2+ concentration. you had to read carefully to notice that haha I missed it too at first

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[@DENNISKIPKOECH-k7j]

check on example 4 and example 7 , you didnt multiply by 2 in example 4 as oppossed to example 7 ( one is reverse of another one )

[@raphmttw3921]

I did the math for Ex.2 and got an answer of 6.476x10^-37. Did you get the same answer?

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4:30 hey guys, he pretty much meant 6.875 x 10^-3 mol Ca(OH)2 when he wrote 6.875 mol Ca(OH)2 right?

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