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Chemistry: Post your doubts here!

sidbloom1995

sidbloom1995

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sitooon said:
its q. 3d
Can you also solve that question of pka of on2006 when you knew the year :D
Click to expand...
i was able to understand it but i wasnt able to solve it (the third part) :p ~sorry; was not much help~
(i) they asked us the pH, for that we will dissociate the acid and form an equation
CH3CH2COOH -----------> CH3CH2COO- + H+
we use this equation in in the formula Ka=[CH3CH2COO-]x[H+]/[CH3CH2COOH]
we have the conc. for [CH3CH2COOH] = 0.5 and Ka = 1.34 x 10^-5
since we have only one H+ and one rest of the molecule CH3CH2COO-, we can assume that they will have equal concentrations! so we will take them as "x"
equating everything:

1.34 x 10^-5 = [x][x]/[0.5]
1.34 x 10^-5 x [0.5] = [x]^2
take square root on both sides and you will get
[x] = 2.6 x 10^-3

now use this conc. in the equation pH= -log[H+]
you will get 2.6 pH

(ii) our equation will be:
NaOH + CH3CH2COOH ---------> [CH3CH2COO-Na+] + H2O
1 : 1 ----------> 1 : 1
0.03 : 0.05 ----------> 0.03 : -
~this part is not asked, it's related to the next part~

(iii) they first found out the no. of moles of the acid, by using it's concentration and volume, we got, 0.5 x 0.1 = 0.05 mol
according to our equation the mole ratio between them is of [1:1] and they already gave us the moles of NaOH as 0.03 mol
0.05 was the no. of moles present in our solution and the no. that reacted with the NaOH was 0.03 (since our acid:base was 1:1) so we subtract them, 0.05-0.03 = 0.02 mol
0.02 mol is the amount of acid left and the 0.03 mol is used up to make sodium propanoate
in this question the concept of excess moles is applied!! the acid has to be in excess or it will all be used up and neutralized, base on the other hand will be all used up will be 100% converted to salt, sodium prppanoate, thus our no. of moles for NaOH = no. of moles of sodium proponoate
thus our sodium proponate is 0.03 mol ~do you get it or is it starting to get confusing?~

(iv) in this part we use two equations, pKa = -logKa -----> eq.1 and the equation for the pH of buffer
pH = pKa + log([salt]/[acid])------> eq.2.................||in case of a base buffer the equation changes pH = (Kw - pKb) + log([base]/[salt]), ours is an acid buffer so we dont need this one right now||
use the eq.1 to find pKa = -log[ 1.34 x 10^-5] = 4.87,
equate it in eq.2
pH = 4.87 + log([0.03]/[0.02]) = 5.046

that will be all
as for that previous qustion i solved it just recently so i'll try to look for it again and tell you when i find the ans! inshallah!
 
S

sitooon

Messages
302
Reaction score
432
Points
28
sidbloom1995 said:
i was able to understand it but i wasnt able to solve it (the third part) :p ~sorry; was not much help~
(i) they asked us the pH, for that we will dissociate the acid and form an equation
CH3CH2COOH -----------> CH3CH2COO- + H+
we use this equation in in the formula Ka=[CH3CH2COO-]x[H+]/[CH3CH2COOH]
we have the conc. for [CH3CH2COOH] = 0.5 and Ka = 1.34 x 10^-5
since we have only one H+ and one rest of the molecule CH3CH2COO-, we can assume that they will have equal concentrations! so we will take them as "x"
equating everything:

1.34 x 10^-5 = [x][x]/[0.5]
1.34 x 10^-5 x [0.5] = [x]^2
take square root on both sides and you will get
[x] = 2.6 x 10^-3

now use this conc. in the equation pH= -log[H+]
you will get 2.6 pH

(ii) our equation will be:
NaOH + CH3CH2COOH ---------> [CH3CH2COO-Na+] + H2O
1 : 1 ----------> 1 : 1
0.03 : 0.05 ----------> 0.03 : -
~this part is not asked, it's related to the next part~

(iii) they first found out the no. of moles of the acid, by using it's concentration and volume, we got, 0.5 x 0.1 = 0.05 mol
according to our equation the mole ratio between them is of [1:1] and they already gave us the moles of NaOH as 0.03 mol
0.05 was the no. of moles present in our solution and the no. that reacted with the NaOH was 0.03 (since our acid:base was 1:1) so we subtract them, 0.05-0.03 = 0.02 mol
0.02 mol is the amount of acid left and the 0.03 mol is used up to make sodium propanoate
in this question the concept of excess moles is applied!! the acid has to be in excess or it will all be used up and neutralized, base on the other hand will be all used up will be 100% converted to salt, sodium prppanoate, thus our no. of moles for NaOH = no. of moles of sodium proponoate
thus our sodium proponate is 0.03 mol ~do you get it or is it starting to get confusing?~

(iv) in this part we use two equations, pKa = -logKa -----> eq.1 and the equation for the pH of buffer
pH = pKa + log([salt]/[acid])------> eq.2.................||in case of a base buffer the equation changes pH = (Kw - pKb) + log([base]/[salt]), ours is an acid buffer so we dont need this one right now||
use the eq.1 to find pKa = -log[ 1.34 x 10^-5] = 4.87,
equate it in eq.2
pH = 4.87 + log([0.03]/[0.02]) = 5.046

that will be all
as for that previous qustion i solved it just recently so i'll try to look for it again and tell you when i find the ans! inshallah!
Click to expand...

May allah shine you with all A*'s for your help to others (y)
A little more explaination of how NaOH and the salt concentration are equal will be appreciated
 
Last edited:
sidbloom1995

sidbloom1995

Messages
83
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114
Points
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i checked your question again!
are you asking about the smooth curves?
just as the names suggest they are curves that are smooth, we usually use then in reaction kinetics most often!
was this what you were asking?
 
sidbloom1995

sidbloom1995

Messages
83
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sitooon said:
The question from on2006 about pka :cool:
Click to expand...
oh ok!
that questuion was asked by someone else and a specific part so i didnt do the pKa part before!
(i) the Ka expression
Ka = [H+][R-]/
............the R is the rest of the group or structure!....in the MS they wrote A but that was confusing because the previous question asked for a compound name A, so i used R instead! ( R is more commonly used to represent groups or structure that are not involved!)

(ii) the conc. of Ibuprofen is 0.15 mol/dm3 and Ka = 6.3 x 10^-6
since ibuprofen have only one carboxylic acid group, (unlike A which have 2)
we know that they divide to form 1 [H+] and 1 [R-], so we know that they have equal conc. and we can donate both of the as "x"

6.3 x 10^-6 = [x][x]/[0.15]
[x]^2 = [6.3 x 10^-6] x [0.15]
take square root on both sides
x = 9.72 x 10^-4
now use the formula pH= -log[H+]
pH= -log[9.72 x 10^-4]
pH= 3.01
 
S

sitooon

Messages
302
Reaction score
432
Points
28
sidbloom1995 said:
oh ok!
that questuion was asked by someone else and a specific part so i didnt do the pKa part before!
(i) the Ka expression
Ka = [H+][R-]/
............the R is the rest of the group or structure!....in the MS they wrote A but that was confusing because the previous question asked for a compound name A, so i used R instead! ( R is more commonly used to represent groups or structure that are not involved!)

(ii) the conc. of Ibuprofen is 0.15 mol/dm3 and Ka = 6.3 x 10^-6
since ibuprofen have only one carboxylic acid group, (unlike A which have 2)
we know that they divide to form 1 [H+] and 1 [R-], so we know that they have equal conc. and we can donate both of the as "x"
6.3 x 10^-6 = [x][x]/[0.15]
[x]^2 = [6.3 x 10^-6] x [0.15]
take square root on both sides
x = 9.72 x 10^-4
now use the formula pH= -log[H+]
pH= -log[9.72 x 10^-4]
pH= 3.01
Click to expand...

I am very sorry , i meant the LAST part !!:(
 
midha.ch

midha.ch

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Someone please help me with Inorganic Chemistry! I need good easy to understand notes!! :(
I would really be hankful if someone helps me! Need it asap
 
S

sitooon

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midha.ch said:
Someone please help me with Inorganic Chemistry! I need good easy to understand notes!! :(
I would really be hankful if someone helps me! Need it asap
Click to expand...
 

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ZaqZainab

ZaqZainab

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midha.ch said:
Thanks :)
I needed notes for organic too
Thanks a looot!! :)
But do you have for inorganic?? You know those transition metals and periodicity and stuffs?:(:(
Click to expand...
I have the notes for all the topics but inorganic
one of notes said
Topics that only contain interactive questions
These topics only contain interactive questions such as animations, multiple choice or audio files. To interact with this content, please go to www.s-cool.co.uk/biology.
 Aliphatic Compounds
 Aromatic and Plastics
 Chemical Energetics
 General Principles
 Ionic Equilibria
 Periodicity
 Transition Metals
do you want this notes they have questions for Group II and Group IV. and the answers
if so here and if no ignore this post
 

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