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Chemistry: Post your doubts here!

S

sitooon

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sidbloom1995 said:
in this one, H2PO4- ats like a base and donated electron and HPO4-2 acts as an acid
the trick is that when a buffer reacts with an acid or a base, the product include the reactent of the other (this is some thing that i have noticed and is not a given rule, watch in the example)

H+ + HPO4-2 ----------> H2PO4-
now
OH- + H2PO4- ---------> HPO4-2 + H2O

as for the part (iii)
you need to use the formula for acidic buffer solutions to find it's pH, which is
pH = pKa + log ( [salt]/[acid])
for pKa u use the value of 7.2, (sorry about this part im a little confused as to how to explain it, form what i remember, i took out the no. of moles of [H+] and used them again in pKa= (-log[H=]) equation and got 7.2 as answer so that's why you have to use 7.2, thought im still not 100% on that, sorry,)
can you give me the paper no. or this, i'll try solving it once more and see if i can answer it
Click to expand...

Well explaind ,
can you help me with this to of finding ( conc. ) of both ?
one other question : whats a Smooth curve ?
upload_2014-3-27_11-41-13.png
 
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S

sitooon

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sidbloom1995 said:
about the acid or base part we have to see if the molecule is giving protons or accepting protons, H+, so it have to do with bronsted lowery base and acid definitions!!
HPO4-2 have only 1 H and can accept one more (since we have H2PO4-) so it will act as a base and accept one more H+ to become H2PO4-.
since there are two salts in our sample, that means that they both exist in our sample. so when an acid is added, HPO4-2 reacts with that acid and accept it's proton H+ and when a base is added, H2PO4- reacts with is and donate is its H+ proton!!
and your paper no. was not right!!
i checked both papers but the question is not there!!
Click to expand...
There are values which are missing !
like pka and others!
she didnt quote it .. its from on2006
 
yousef

yousef

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When add NaOH with compound with has C=C bond &carboxilic acid ( eg. propenoic acid ).. will it be addition / condensation
and do we consider NaOH as functional group ??
 
ZIYAD.beasto

ZIYAD.beasto

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XPFMember said:
Hi everyone, AsSalamoAlaikum Wr Wb..

To get things organized in a better way, I am making this thread. As othewise, some queries remain unanswered!

So post your CHEMISTRY doubts in this thread. InshaAllah members around will help you.

Any Chemistry related notes and links will be added here in this post. Feel free to provide the links to your notes around the forum, or any other websites!

Chemistry Notes:


http://www.chemguide.co.uk This is the website, which contains almost everything classified according to the syllabus.

Tips for solving chemistry MCQs 9701/01

Chemistry P5 Tips and Notes

Some links & Notes - by 'destined007'

Chemistry worksheets Link shared by hassam

Chemistry Application Booklet: Mistakes and Corrections!

Calculations for A level Chemistry, author E.N. Ramsden third edition ebook download.


Regards,
XPC Staff.
Click to expand...
XPFMember said:
Hi everyone, AsSalamoAlaikum Wr Wb..

To get things organized in a better way, I am making this thread. As othewise, some queries remain unanswered!

So post your CHEMISTRY doubts in this thread. InshaAllah members around will help you.

Any Chemistry related notes and links will be added here in this post. Feel free to provide the links to your notes around the forum, or any other websites!

Chemistry Notes:


http://www.chemguide.co.uk This is the website, which contains almost everything classified according to the syllabus.

Tips for solving chemistry MCQs 9701/01

Chemistry P5 Tips and Notes

Some links & Notes - by 'destined007'

Chemistry worksheets Link shared by hassam

Chemistry Application Booklet: Mistakes and Corrections!

Calculations for A level Chemistry, author E.N. Ramsden third edition ebook download.


Regards,
XPC Staff.
Click to expand...
hello please need help with Chemistry A2 P5 Q2 solved
 
F

Faaiz Haque

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ZaqZainab said:
The general formula for ester is there in the pic It is always suppose to be there in an ester if it isn't then that is not an ester View attachment 37732
So they have given me C5H10O2

CH3-CH2-CH2-CH2-COOH----->C4H9-CO2H

CH3-CH2-CH2-COO-CH3------->C3H7CO2CH3

CH3-CH2-COO-CH2-CH3------>C2H5CO2C2H5

I am just moving the COO group If you notice in the first one the COO group is in the end and then in second one its in the second last place and in the third one its is in the middle You have to draw this with all the Bonds showing
Click to expand...
ty
 
sidbloom1995

sidbloom1995

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yousef said:
When add NaOH with compound with has C=C bond &carboxilic acid ( eg. propenoic acid ).. will it be addition / condensation
and do we consider NaOH as functional group ??
Click to expand...
i dont know of a reaction between >C=C< with NaOH but with COO2H the H goes to the -OH of NaOH and form H2O and the becomes CO2-Na+ it will be a dehydration reaction!
and no NaOH is not a functional group!! ( i still don get why you said that but so far that will be the ans. to it)
 
sidbloom1995

sidbloom1995

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1357911 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_4.pdf
Q-2 a iv?? Can someone please help me??:unsure:
Click to expand...
the question states that it is a dibasic acid, that means that there are two -COOH groups present at it's two ends
you first find the no. of moles of the NaOH, which in this case is 1.2 x 10^-3
then since you know that there are two acidic groups then that mean that we will need 2 molecules of NaOH for 1 molecule of "A"
so we use ratio method to deduce the moles of "A" which will be [1.2 x 10^-3]/2 = 6 x 10^-4
now you have the moles and the question have given you the mass of 0.1g so use that to get the Mr., [0.1]/[6 x 10^-4] = 166.6 which we will round off to 166
we already now that we have 2[-COOH] so that is a must add and it's Mr. is 90! (you can calculate that)
subtract 90 from 166 and you will get 76!
that means that your rest of the mass of C and H should add up to 76
in this case we will have 3C and 4H to add up to 76 so that will be the rest of your structure, giving you some thing like this: HOOC-C6H4--COOH
 
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