Titration of a strong acid with a strong base | Chemistry | Khan Academy

Calculating the pH before the equivalence point for titration of strong acid, hydrochloric acid, with strong base, NaOH. Created by Jay.
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[@user-fy9di3yo5y]

i can pretend i understand it

[@Hakseng127]

plz mention end point, indicators, buffer region, nd equivalance point too

[@icassipops]

that's the very first "not that helpful" "complex" khan academy video i've ever watched but anyway thanks for the effort

[@mbtisocialclub]

This video is perfect . Very well explained and right to the point . Easy to understand compared to my chem prof ...

[@b-b-ntheboy5613]

Thank you very much it was exactly what i was looking for

[@nurlybektoktarov469]

Lifehack: put it at the speed 1.5 and Khan Academy will speak like normal human 😄

[@Chirp-chirp]

Really? I feel like he speaks so much faster in the chem videos compared to the math ones

[@zidamv1944]

very well explained sir ! thank you

[@terrencegreen755]

thank you!

[@ran9628]

Thanks!

[@pasha1339]

This man is the best. Thank youuuuuuuuuuuuuu

[@khattakfarhan9854]

thank you so much sir

[@priyankapal5599]

Thank u so much sir

[@OctobrTea]

thank you soooooooooooooooooooooooooooooooooooooooooooooooooo much

[@chamyke]

Thanks for the video it is very useful:) I have an other (for me easier ) way to find the concentration of H3O+.
You have all in all 20 ml of HCl , it will be added 10 ml of NaOH which is 1/2 of the whole amount.
So you can say: [H3O+] = 0,5*1/2* 20ml/30ml = 0,167
PH = -log (0,167)= 0,777

[@kimmadarimot9868]

hello just want to ask. Is the hydronium ion concentration of a strong acid titrated with a strong base equal to the concentration of the strong base at the equivalent point?

[@sabaakbar7427]

At end point/ eq.point conecentration of hydronium ion is more than conetration of base in that case that is why solution is acidic in nature because all base moles have been neutralized with moles of acid but acid moles are more than base

[@emilys.8347]

Is it possible to titrate a strong base (analyte) with a strong acid (titrant)??

[@peybak]

Yes, the curve will be the opposite of this one, meaning that the starting pH will be high.

[@dayanacheong901]

can the calculation of the titration of strong base and strong acid use the handerson hasselbalch eqn?

[@jakobj58]

no. you can only use the henderson.hasselbalch equation for weak acids because every molecule of the acid is deprotonated. the equation would then be: ph= pks - log(0/A-)
-> ph = pks - log (0). log (0) is not defined

[@atreyobhatta9919]

Nina Dobrev that's the formula for finding pH

[@user-uh9uv2yw9h]

No. The Henderson eq is used ONLY for buffers

[@filipdepay4155]

pozdrawiam 8 grupe !

[@laraibkhan9102]

anybody help me in solving some chemistry problems ASAP ??

[@esa2236]

Thanks so much for the video, but I think all the extra explanation like converting ml to L is very excessive if this video was made for students in organic chemistry....

[@TheRelentlessPPL]

Skid some do this in high school so let it be Mr. Chemistry

[@brainphelps1994]

no skid is right if you're doing this in high school and don't know how to convert moles and mL and all that crap you are the anomaly. The conversions are tedious

[@kristinmcelroy5356]

how do i find Ka from this

[@benhardsim8629]

i know i am 2 years late , but maybe its gonna be useful for someone else .
you cannot find the Ka because this is a strong acid and base solution .

[@abdulmunimjundurahman7923]

@@benhardsim8629 After 1 years you helped someone with the same question.

[@stoneluo8986]

why cant you just get khan to explain ...

[@arbindapandey1711]

I didn't understood why the concentration of H3O+ was also 0.500M.

[@beneastman9244]

Strong acids completely dissociate in aq solutions. So, the concentration of H3O+ is going to equal whatever the concentration of HCl is because there is no base added yet.

[@beneastman9244]

In other words, every single H+ that was once bound to Cl- is now bound to H2O.

[@rustyshackelford3590]

The MCAT can go to hell for making us do this without a calculator.

[@khattakfarhan9854]

thank you so much sir

[@khattakfarhan9854]

thank you so much sir