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Chemistry: Post your doubts here!

V

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Calculate the pH of the buffer formed when 10.0 cm3 of 0.100 mol dm–3 NaOH is added to 10.0 cm3 of 0.250 mol dm–3 CH3 CO2H, whose pKa
= 4.76.
 
knowitall10

knowitall10

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veteran said:
Calculate the pH of the buffer formed when 10.0 cm3 of 0.100 mol dm–3 NaOH is added to 10.0 cm3 of 0.250 mol dm–3 CH3 CO2H, whose pKa
= 4.76.
Click to expand...
That's the same question the brother posted right before you:) Can you please tell me the answer? I'll try to work backwards.
 
knowitall10

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veteran said:
Yeah but what i dont get is how he came up with those values, can you give be a step by step solution? i would really appreciate ir.
Click to expand...
Ok...help me out with this alright? Let's work it together:
What i get is that it will form the salt sodium ethanoate....what next?
 
V

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knowitall10 said:
Ok...help me out with this alright? Let's work it together:
What i get is that it will form the salt sodium ethanoate....what next?
Click to expand...
i think i got it. I calculated the equilibrium concentration between the acid and the salt being produced. I assumed that the acid will produce only half of the CH3COO- ions so moles =10/1000 x 0.250 and divide the answer by 2 to get the moles of the ethanoate ion. Then do the same for NaOH moles 10/1000 x 0.1.. and divide the answer by two to get the value of sodium ions. That is the same as the concentration of sodium ethanoate. Then get concetration of the salt. Subtract the moles of salt from that of the moles of ethanoate ions and then use that formula. I hope that makes some sense.
 
littlecloud11

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raihan1904 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_4.pdf
Q2 part (b) explanation plzz
plzz tag me in the explanation of this question!!:)
thanks
best ov luck evry1
Click to expand...

Here-
iKhaled explained this earlier.
C40H82 -------> C16H34 + C24H48

a good way of explaining this question is to see how many bonds u have before the reaction and how many bonds u have after the reaction for example...

before the reaction we have 39 C-C bonds and 82 C-H bonds...after the reaction we have 15+22 = 37 C-C bonds and one C=C bond and 48+34 = 82 C-H bonds

so we have broken 39-37 = 2 C-C bonds and formed 1 C=C bond
bond broken = 2 x 350 = +700
bond formed = -610 = -610
enthalpy = + 90 kjmol^-1 !!

thats it..the mark scheme got the answer as +180 because he has used a different equation in cracking instead of forming C24H48 as our alkene he formed 2 moles of C12H24 and in that way we will have 2 C=C bonds instead of one and 4 C-C bonds instead of 2 because one mole of C12H24 contains 1 C=C bond and 10 C-C bonds so the 2 moles will contain 20 c-c bond and 2 C=C bonds 20+15= 35 so 39-35 = 4 C-C bonds so we had broken 4 C-C bonds and formed 2 C=C bond..in that way u will get 180 kj per mole instead of 90..both are correct though depending on the equation u made ;)

i hope u got it!!
Click to expand...
 
knowitall10

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veteran said:
i think i got it. I calculated the equilibrium concentration between the acid and the salt being produced. I assumed that the acid will produce only half of the CH3COO- ions so moles =10/1000 x 0.250 and divide the answer by 2 to get the moles of the ethanoate ion. Then do the same for NaOH moles 10/1000 x 0.1.. and divide the answer by two to get the value of sodium ions. That is the same as the concentration of sodium ethanoate. Then get concetration of the salt. Subtract the moles of salt from that of the moles of ethanoate ions and then use that formula. I hope that makes some sense.
Click to expand...
Yeah...it makes sense.... but why do we divide by 2? But (y) That's one decent explanation!
 
Xtremite

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knowitall10 said:
Ring Activating groups (Substitute in the 2,4 and 6 positions): -alkyl, -OH, -NH3
Ring De-activating groups (substitute at the 3 and 5 positions): -COOH, -NO2, -carbonyl, -Cl2
Click to expand...

I think it'd be better if we call the Ring Activating group ' Electron donating group' and the Ring Deactivating one ' Electron Withdrawing group'.

Thanks for the points. :)
 
knowitall10

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Xtremite said:
I think it'd be better if we call the Ring Activating group ' Electron donating group' and the Ring Deactivating one ' Electron Withdrawing group'.

Thanks for the points. :)
Click to expand...
Yeah... we can call it anything. But Electron donating and Electron withdrawing is better. ur right:)
 
knowitall10

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Xtremite said:
I think it'd be better if we call the Ring Activating group ' Electron donating group' and the Ring Deactivating one ' Electron Withdrawing group'.

Thanks for the points. :)
Click to expand...
and...:) Don't mention it...c'mon, we're all here to help:):) So u dnt need to thank me, just keep me in your prayers...
 
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