Practice Problem: Calculations Involving pH and Ka

[@shebisdead]

your videos are the only thing getting me through college

[@brianchanda2594]

I love the content profesor

[@zencodm9873]

thank you chemistry jesus

[@theyaghu]

It is useful to me

[@myst1317]

very good

[@Eva-xy9zd]

How do I calculate both concentration and pH of all molecules and ions, if I'm only given that a propinoic acid in an aqueous solution has a Ka = 1,3*10^-5 and a concention on 0,100M ? I've made an ice box, but the process kind of stops then..

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[@daxtaylor9514]

@Kyrie Dustin Instablaster ;)

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@Dax Taylor i really appreciate your reply. I got to the site thru google and Im in the hacking process now. Seems to take quite some time so I will reply here later when my account password hopefully is recovered.

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@Dax Taylor It worked and I now got access to my account again. I'm so happy! Thank you so much, you saved my account!

[@daxtaylor9514]

@Kyrie Dustin happy to help :)

[@fhjebbb6382]

Could have done with this before my chemistry alevel exam ://

[@seikomyazawa]

Profesor Dave, I've used the formula pH= 1/2(pka - log(acide concentration) and solved for pka, I've got pka = 2pH + log(acide concentration) and then to get the ka I've just raised 10 to the negative pka, but I've got ka= 4.05*10^-4 instead of 4.5*10^-4 and I can't find out where I made a mistake, can you explaine please

[@berserk4447]

I'm not sure what the mistake is but that isn't the equation I recommend you use to solve the problem. take 10^-pH to find the hydronium molarity which will be the same as the conjugate base molarity. Then divide it by the quantity of the initial acid molarity - the hydronium molarity. Now I see this was from two years ago. rip. oh well

[@seikomyazawa]

@@berserk4447 still alive lol. Thanks for your answer

[@spicydiarrhea5662]

@@seikomyazawa lol just wanted to ask the exact same question as you! I also got ka= 4.05*10^-4 initially, but then used the proper full formula K=(C*alpha^2 ) / 1-alpha and got 4,4425586*10^-4. (for alpha >5% full formula is a must) Mine result is more accurate than professor's Dave - exact even, worked on proper calculator without any approximations

[@noorfatima0699]

YESSS! i also got this answer by using the initial concentration of the HNO2 as 0.0516 instead of using the equilibrium concentration which in the video he says is 0.047. I think the answers are pretty similar so it should be good enough?

[@Angelusloco15]

First comment! Thanksgiving you for explaining this Prof

[@RabanyMedo]

Every time he says "icebox" I die inside lol.

[@itzjowii]

I’ll force feed you pine cones

[@karthikroddom]

Hi sir, is it possible for you to explain acidic and basic buffers and some problems related to it as well. ~Thanks

[@ProfessorDaveExplains]

i've got a tutorial on buffers, but i think there are a couple problems coming up as well

[@Animalia.5]

I'm confused with 10 in last equation

[@jamesprince571]

Inverse of log is 10 to power of

[@tanyashrivastava1726]

Hey, I need to calculate how much the quantity of Conc H2SO4 Will be required for a 100 mL solution to reduce the pH from 6.5 to 4.5. How do I calculate it ?? Thank you so much in advance

[@godofplatinum124]

Legend

[@JustShamiaBaby]

Where is the ice box tutorial video?

[@ProfessorDaveExplains]

in the general chemistry playlist

[@punisher0717]

If you can’t do… teach i suppose

[@maxthilen5434]

gotta mute the intro