Molar Solubility Example Calculation - Common Ion

This short video is a worked example of calculating the molar solubility of an insoluble compound in the presence of a common ion.

[@roye6961]

you explained it better than many other videos in youtube, i had a fk headache asking why this and why that on other videos solving similar problems, but you left no questions

[@elinaoberemok1732]

Thank you! This was so easy to understand with your video :)!

[@Youcanbeeu]

You've saved my brains, THANK YOU!

[@MrBlash93]

Thank you very much for making this video. You really have helped me. Thank you

[@stephenjohnston9708]

Thank you, perfect explanation!

[@mohamedamin63217]

شكرا جدا

[@koweteaches961]

sorry pls what does ksp mean

[@MadLifeTurtle123]

solubility product

[@rugrat123abc1995]

Thanks so much!

[@samyentertainment3086]

i'm pretty sure you forgot to square x :)

[@hifriede8581]

Nah you don't need to

[@Mggames-ds5dm]

you don't need to because none of the ions have a coefficient of two

[@johnspence8141]

Try it both ways (x^2 + 0.5x - 4.2*10-8) = 0, and the solutions for x are the same as 0.5x= 4.2 *10-8, thats because the 0.5M of bromine already present is WAYYYYY off the charts compares to the bromine added, so its really the 0.5M of bromine that is doubling the mols/L.

[@diegobravo3870]

The common ion effect

[@TurmyHD]

You didn't square root at the end, so your answer is wrong. It's 2.9x10 -4

[@dianaperez63]

Nope, She's right actually. I was confused at first too but she is right.

[@hifriede8581]

She's right you don't need to

[@gomez2858]

@@hifriede8581 why tho

[@smaccsi]

wdym why tho, the f u gonna square root for ?

[@johnspence8141]

@@gomez2858 Try it both ways (x^2 + 0.5x - 4.2*10-8) = 0, and the solutions for x are the same as 0.5x= 4.2 *10-8, thats because the 0.5M of bromine already present is WAYYYYY off the charts compares to the bromine added, so its really the 0.5M of bromine that is doubling the mols/L.

[@waynenelson9095]

What if it was 2:1

[@johnspence8141]

Thats why you do the chemical equation, to determine how many MOLs are in each side of the equation, and you adjust accordingly based on what you find...but beware if its twice Molarity it will also affect the solubility constant equation, as well as the final MOL amount.