AQA A-Level Chemistry - Kw and Bases

This video explains the term Kw and how it can be used to find the pH of bases.

The notes are available here:

[oliviabirkett]

Legend has it he’s still looking for the perfect salmon colour

[black_hammer]

six years later and you are still a blessing to hundreds of students, we really do appreciate your effort!! : )

[munig]

"that's a lovely colour isn't it... like skin!" love this

[xencha]

"that's a lovely colour isn't it" hahahaha

[melissapool]

“Possibly dribbling out your eyes” omg made my day

[talabashar]

8 years later.. still helping us out what a G

[Carlos-vdni]

Is there a video on Ka and working out weak acids and bases?

[BeastlyProduction]

Thank you! Youve helped me at AS, and with the new content continue to help me at A2 :).

[weshouldsaveourselves]

god bless your soul. You are so genuine and very kind. You deserve more subs dude

[daniellaruk]

😂 the salmon hahaha, honestly made me laugh so much through the stress of paper 1 tomorrow

[e.b]

I am happy you have uploaded videos on this topic, thank you so so much!

[jessm]

could you please make a video that covers optical isomerism?

[thewarhammerworkshop]

On the calculations at around 11:50 how were you getting the values for KW or were you just not showing them to us?

[reiantaank]

Thank You so much. Your videos are amazing! Can you do a quick video on pH curves please?

[The_Reductionist]

Thanks for the new A2 vids, much appreciated

[thatskykidfreya]

I’m confused at 11:00
I put the equation into my calculator and got 3.7 not 13.70.
I did:
1x10^-4 / 0.500 = 2x10^-4

pH= -log ( 2x10^-4 ) = 3.7

HELP!!

[mouchal-zubaydi]

This is amazing! Will you upload a video on indicators ?

[NiksDjs]

Hi!,
How does temperature effect the position of equilibrium of kw? Is it in warmer temperatures that water is more 'acidic' ?
The video is fantastic by the way! thank you!

[MH-ytvm]

How did you get the answer of 14.79 at 11:59?

[uzman]

Hello! Loved the video...really helped, but I’m a little confused. At 7:04, how did you get a pH of 6.14? I tried it and I got something much different.

Method:
Kw= 51.3x10^-14
Kw= [H+][OH-]
Kw= 2[H+]
[H+]= (51.3x10^-14)/2
[H+]= 2.565x10^-13
pH=-log[H+]
pH=-log[2.565x10^-13]
pH=
Please help. Thanks