Solubility Curves - Basic Introduction - Chemistry Problems

This chemistry video tutorial provides a basic introduction into solubility curves. It contains plenty of examples and practice problems on calculating the solubility of an ionic compound at a given temperature and volume. It explains how to determine the temperature at which two substances have the same solubility in addition to determining if a solution is saturated, unsaturated, or supersaturated under certain conditions.

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Heating Curve of Water:

Heating Curve Chemistry Problems:

Final Temperature - Ice Water Mixture:

Molarity, Molality, Density, & Mass %:

Normality & Equivalent Weight:

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PPM and PPB Concentrations:

How To Convert PPM to Molarity:

Enthalpy of Solution & Hydration:

Solubility Vs Concentration:

Solubility Curves:

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Henry's Law & Gas Solubility:

Vapor Pressure & Clausius Equation:

Raoult's Law - Vapor Pressure:

Colligative Properties:

Chemical Kinetics Initial Rate Method:

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Final Exams and Video Playlists:

Full-Length Videos and Worksheets:

Chemistry PDF Worksheets:

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Look I’m not going to lie I sat down in class and nearly screamed because this confused me so much but literally watching this video made it click to me. I swear your a genius and your videos are very descriptive. Please continue to make more chemistry videos and help educate the confused

[justanotherweirdo]

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[georgesadler]

Professor Organic Chemistry Tutor, thank you for a Graphical and Analytical explanation of Solubility Curves in AP/General Chemistry. Solving practice problems using graphical tools /methods is an excellent way to visualize and understand the material. This is an error free video/lecture on YouTube TV with the Organic Chemistry Tutor.

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[judeekiribi_dbassist]

Thanks for the lecture. It was quite interesting and easy to understand graphically. However, I want to also confirm if the answer can be 62.5g/100ml.

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at 15.43, I think we don't have a supersaturated solution .we have a solution with undissolved particles thanks for video btw.

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[papaamponn-nyamekye]

Hi, with Q7 (d), I multiplied the solubility of 50 g of KNO3 in 100 mL at 30 degrees C, by 0.25 (1/4) in order to get it to the same value of Volume. This resulted in a solubility of 12.5 g in 25 mL. This meant that given that the Conc. of 15 g of KNO3 in 25 mL was greater than (>) the solubility of 12.5 g of KNO3 in 25 mL. Thus C > S, therefore arriving at the same answer of super saturated. Now given that in the video you calculated to make both conc. and solubility in terms of 100 mL, which makes finding the answer on the solubility curve more of a possiblity.

I am wondering if my method is acceptable if I was simply comparing conc. to solubility and not looking at the solubility curve?

Thank you,
Papa

[denn_ber]

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