16.5 pH Calculations for Weak Acids and Bases | General Chemistry

[@yesenianieto9259]

I think the Kb value should be around 2.5e-7 for the last question. Multiple people have commented the same answer vs his answer.

[@QuicksilverBL3DE]

Just checked what Chad did he plugged the values as follows (10^-3.8^2)/(0.1-10^-3.8), you see what he did wrong the correct way is (10^-3.8)^2/(0.1-10^-3.8)

[@ChristianSpencer-c9g]

I agree, when Chat GPT and Chad gets the wrong answer you know your in to deep.

[@StephaniFernando-ml8qx]

Yes, that's what I've got too

[@user-ux6qb4xv2o]

I cannot express how amazing all of your videos are! I went from barely scraping by to getting a 90% on my last exam. :) Thank you ❤

[@ChadsPrep]

Great to hear!

[@Brookieats210]

I dont get why your videos don't have more thumbs ups and views. You have saved me this semester in Gen Chem 2! Thanks Chad!

[@ChadsPrep]

You're welcome, Brooke - glad the channel is helping so much!

[@courtneyjohnson5529]

I tutor for general chemistry and I mainly use your videos as a reference! God bless your channel and family!!

[@ChadsPrep]

Excellent and Thank You!

[@SuperJustachannel]

I truly appreciate your channel and all the work you put into making concepts easily digestible. You obviously love what you do and you're phenomenal at it! Much love and appreciation

[@ChadsPrep]

Thanks for saying so!

[@BobDaMan-e4w]

greatest youtube teacher all time

[@ChadsPrep]

Wow - that's high praise - thank you!

[@triple_gem_shining]

kb on last example is 2.5x10^-7 to confirm for those confused. :)

[@triple_gem_shining]

we all make mistakes.

[@Actiaeon]

Thank you, I was getting so frustrated with weak acids. But you explained it in a way I completely understand.

[@ChadsPrep]

You're welcome - Glad it became clear!

[@sandiecarson9738]

You videos are so easy to understand, you are the best.

[@ChadsPrep]

Thanks very much.

[@senema718]

I'm happy to find your channel. I study at İzmir Institute of Technology in Türkiye and my department is chemistry. Our lessons' is 100% english. But my english skills is not enough to understand the classes. But your videos helps to me understand very well. (sometimes speed: 0.75x) Thank you.

[@ChadsPrep]

Glad you found the channel and that it is helping you - Happy Studying!

[@cesaro876]

Thank you very much Chad, I love how simple you make everything man ❤

[@ChadsPrep]

You're welcome and Thank you.

[@lilspacebunnny]

Thank you so much for this good explanation!!! I finally understand it

[@ChadsPrep]

Excellent!

[@QFBPLR]

Hola, Chad, sólo te escribo para agradecerte por tus vídeos, porque, gracias a eso, fuí aceptado en la maestría. Espero te encuentres bien y te deseo todo el éxito y la salud del mundo. Te envío un abrazo desde México.

[@ChadsPrep]

Awesome! Congratulations! And you are most welcome! And best wishes in the masters program and beyond! 🎓

[@00-1B]

Another short way of finding pH of weak acids is: "pH=1/2 (pKa - log[HA]i)". Following this method I got pH of 2.875 (round up to 2.9) for the first example. The same thing goes for pOH; substituting pKa for pKb and [HA]i for [B]i respectively. Just a method I was taught, I thought it would be interesting to share. Love your videos, Chad! You helped me immensely

[@phonognome6218]

yup he's got rounding errors.

[@jernejkolar5028]

absolute chad

[@WisaniHlungwani]

GAINED A SUB BECAUSE OF THAT SHIRT CHAD. [ THANKS FOR THIS PRODUCTIVE LESSON]😍

[@ChadsPrep]

Thank you and welcome to the channel!

[@jenniferyoung9472]

Your videos are awesome, and I'm so grateful for you help! Just a note though, I ran the numbers several times (even just using the decimals instead of scientific notation) and kept getting 1.78 x 10^-5 for Kb. Is there a chance your calculator didn't format your entry correctly on the last problem in the video?

[@Sokreah]

Bro living up to his name.

[@ChadsPrep]

Glad you like the channel!

[@ThuNguyen-wr5sg]

Hi Chad. at 35:15 when calculating Kb with that formula I get different answer. (2.5e-7 instead of 3.6e14). Can you help?

[@박서정-r5z]

i also got the same value

[@premp032]

Yea i got the same answer too, I think he may have made a mistake, rare, but i think it might have happend

[@swasgrover3414]

Yeah I got the same answer

[@chrishaddad7983]

me too

[@QuicksilverBL3DE]

Just checked what Chad did he plugged the values as follows (10^-3.8^2)/(0.1-10^-3.8), you see what he did wrong the correct way is (10^-3.8)^2/(0.1-10^-3.8)

[@uNknownrMx]

You’re the best ❤️

[@ChadsPrep]

Thank you

[@glykerkele6651]

14:16 if we had solved the quadratic would we have found the same value for x? If not then would we have found a bigger or a smaller x?

[@pika3856]

saving lives out here

[@ChadsPrep]

Saving grades maybe, but hopefully just saving time!😉

[@yusufyusuf7919]

Explained it way better then my trash chem teacher, u da goat.

[@ChadsPrep]

Glad the channel is helping you!

[@GeorgeCronigen]

Thank you brother. Your help is precious in an Era that good teachers are scarce. I think that the last caclulation of Kb in the last exercise is wrong

[@ChadsPrep]

Glad the channel is helping you!

[@TheRooflesstoofless]

You’re sooooo not a “Chad”. Thanks Chad!!! 🎉

[@ChadsPrep]

You're welcome!

[@thenightdances21]

Hi Chad, I don’t understand the relationship between Ka and Kb and Kw. Why does the Kb have to be smaller than Ka in order for Kw to hold at time 28:15 ? Can you please explain this again?

[@GeorgeCronigen]

Ka multiplied by Kb will always give you kw which equals to 1x10-¹⁴. Kw alone equals to [OH-] x [H+] which exist in the water due to its autoionization. These concentrations mentioned are each of them equal to 10‐⁷, hence 10-⁷ × 10-⁷ = 10-¹⁴. When you pure an acid in the water this balance will tilt in favor of H+ and for example if the end concentration of Η+ after the addition of acid reaches for example 10-¹ the concentration will be reduced from 10‐⁷ to 10-¹³ in order to satisfy this relationship. Think of it this way if you put acid in water you create many protons which eliminate free OH- . I hope I have helped

[@thenightdances21]

@@GeorgeCronigen Hi George! Thank you SO much!! Your explanation helped and makes a lot of sense!! I really appreciate your help and detail response!

[@GeorgeCronigen]

@thenightdances21 all good brother glad that I helped you

[@anaismontoya1803]

Hello Chad! Thank you so much for these videos - they help tremendously! For this one, at around 35min, my calculator keeps displaying an error message each time I try to calculate with 10^-3.8. I've tried other variations of this but continue to get the error message. Are you typing in something different?

[@anaismontoya1803]

My calculator just doesn't want to work with decimals in exponents :/

[@anaismontoya1803]

Figured out my mistake. Didn't realize 10^-3.8 was the inverse of log! :)

[@RelaxRailRide]

Nice teaching way 💕💕

[@ChadsPrep]

Thanks!

[@RelaxRailRide]

Sir, need coordination chemistry, statistical thermodynamics, group theory, and quantum chemistry video s... 😊

[@ChadsPrep]

Thanks for the feedback Chemistry wide - noted :)

[@RelaxRailRide]

@@ChadsPrep ok ok sir.. 💐

[@sivafaraj1342]

you are the best

[@ChadsPrep]

Thank you

[@AbleWogu]

Chad you did a great Job and thanks for exposing me to an easier formula.. but I think the last answer you got was wrong , I know it was out of not going detailed in that question, but I think the answer us Kb= 2.6*10^-7.. I also used your formula to confirm.. and I got the same, so I think u made mistake somewhere 😢😢

[@lilcody69]

When I pluged the values of (10^-3.8)^2/(.1-10^-3.8) my value equaled 2.51 * 10^-7. Where did the 3.6*10^-14 come from, what step am I missing/forgetting? Please and thank you Chad

[@sandomachera]

same. i thought iw as going insane.

[@QuicksilverBL3DE]

Just checked what Chad did he plugged the values as follows (10^-3.8^2)/(0.1-10^-3.8), you see what he did wrong the correct way is (10^-3.8)^2/(0.1-10^-3.8)

[@neaworld3960]

GOD BLESS YOU! THE LORD GOD of Heaven Bless you and keep you, and make His face shine upon you and be gracious unto yiu, and give you His peace, and His countenance.

[@ChadsPrep]

He does every day - and to you as well!