Common ion effect and buffers | Chemistry | Khan Academy 

I swear you and organic chem tutor have the lowest volume videos on youtube.

Y'all *demanding* a solution without calculator should really be ashamed not to appreciate the work this man is doing. If you want to know how to approximate the value of pH, that's not something that the chemistry video must necessarily cover. Go search RU-vid for a video on logarithms.

Check this out. [anything] = n x 10^(m) p[anything] = (n-1).(10-m) So in the video, Ka= 1.8 x 10^(-5) = [H3O+] p[H3O+]= (5-1).(10-2) [don't include negative & round). pH= 4.8 ~4.74 Learned this through a MCAT Kaplan Video, so the credit goes to them! You can use this for anything...so if you're given Ka and need to get the pKa, or the pOH and you know the [OH]. Very handy for fast calculations. You can always to the long-log method for an accurate answer.

Thank you. Really learn more here than lessons.

Wooow thanks Khan Academy. I have always had troubles in this topic. Now am happy I have followed

Henderson Hassleback equation will be your friend with these type problems. PH= Pka+log ([conjugate base]/[acid])

Why is the volume so low?

thank you so much khan academy! this helped a great deal! :-))

10:56 Can we assume every time that the x is of a very less value? THank u for the video!

Do we have to always assume that x is a very small number?

Yaaaas, thank you!

How can we equate Ka = [CH3COO-][H3O+]/[CH3COOH] what I have learnt Is this [CH3COO-][H3O+]/[CH3COOH] = Kh (i.e hydrolysis constant) and Kh = Kw/Ka. And from here we calculate x and then [H3O+] = x. And then pH = - log (x).

thank you, this video was so helpful ! :)

Thank you!! never knew it was this easy

Thank you so much

OK I'm sorry. I was wrong Kh = Kw/Ka is in case of hydrolysis of salt. Thanks. For video

Excellent, thanks!

Thank you...

best explanation

You cannot use a calculator on the MCAT. You should show people how to best approximate these values. Example: -log(m x 10^-n) = approx. n - 0.m. So for this video -log (7.7 x 10^ -6) = 6- 0.77 = 5.23

not me abusing the Henderson-Hasselbalch equation to get the same answers

gless u man saved my grade

Thank you so much sir! I 🥹🥹

What if Ka and Kb are not given? How would we find the pH buffer?

thank youuuuuuuuuuuuuuuuuuuuuuuuuuu

Thx! Much help

thank you

good

thank for sharing

thanks...

Thanks!

whose watching after 6years

All your videos are calculate pH, the EASIEST. Why not do an example on calculate strong base conc. if the pH of acetic acid buffer is _____________.

HKUST anyone?

Why is this not a buffer system ? why didn't we take the -log of the Ka and then the log of the ratio of conjugate base over acid?

why is the x small??

Please fix the sound

Arrigato

What does happen to Nh4NO3??

These are AAMC approved videos and given that use of calculator is not permitted in the exam, is it the time you showed your viewers how to solve these problems WITHOUT a calculator?

Plz also translate ur lectures into urdu........... advance thanks :-)

Why is the volume always so low in khanacademy videos?

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how can pH be greater than 7???

chemistry sucks but thx for the video anyway

Marry me professor

Am i the only one who dint get this lecture??

rubbish

Thank you so so so muchhh , I have my chemistry exam tomorrow ( 6 units included ) & you're being a lifesaver rn. 🤍