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Chemistry: Post your doubts here!

Metanoia

Metanoia

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DeViL gURl B) said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s13_qp_42.pdf

can someone please help me in question 2 part b.
thank you.
Click to expand...

Let volume of CH3COOH be x dm3
volume of CH3COONa = (0.1 - x) dm3

moles of CH3COOH = x mol
moles of CH3COONa = (0.1 - x) mol

[CH3COONa]/[CH3COOH] = x/(0.1 -x)

pH = pKa + lg ([conjugate base]/[acid])
5.5 = - lg (0.0000179) + lg [x/(0.1 -x)]
5.5 = 4.747 + lg [x/(0.1 -x)]
0.75285 = lg [x/(0.1 -x)]

5.6604 = x/(0.1 - x)
0.56604 - 5.6604 x = x
6.6604 x = 0.56604
x = 0.085
 
DeViL gURl B)

DeViL gURl B)

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Metanoia said:
Let volume of CH3COOH be x dm3
volume of CH3COONa = (0.1 - x) dm3

moles of CH3COOH = x mol
moles of CH3COONa = (0.1 - x) mol

[CH3COONa]/[CH3COOH] = x/(0.1 -x)

pH = pKa + lg ([conjugate base]/[acid])
5.5 = - lg (0.0000179) + lg [x/(0.1 -x)]
5.5 = 4.747 + lg [x/(0.1 -x)]
0.75285 = lg [x/(0.1 -x)]

5.6604 = x/(0.1 - x)
0.56604 - 5.6604 x = x
6.6604 x = 0.56604
x = 0.085
Click to expand...

Thank you soo much for the explanation. ^.^
 
DeViL gURl B)

DeViL gURl B)

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Metanoia said:
Let volume of CH3COOH be x dm3
volume of CH3COONa = (0.1 - x) dm3

moles of CH3COOH = x mol
moles of CH3COONa = (0.1 - x) mol

[CH3COONa]/[CH3COOH] = x/(0.1 -x)

pH = pKa + lg ([conjugate base]/[acid])
5.5 = - lg (0.0000179) + lg [x/(0.1 -x)]
5.5 = 4.747 + lg [x/(0.1 -x)]
0.75285 = lg [x/(0.1 -x)]

5.6604 = x/(0.1 - x)
0.56604 - 5.6604 x = x
6.6604 x = 0.56604
x = 0.085
Click to expand...

but can you please tell me why did you use the volume in the brackets, that is the salt and the acid bracket.
cuz what we've been told is that it should've the concentration. .-.
 
Metanoia

Metanoia

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DeViL gURl B) said:
but can you please tell me why did you use the volume in the brackets, that is the salt and the acid bracket.
cuz what we've been told is that it should've the concentration. .-.
Click to expand...

In this case , they are numerically same because the concentrations of both solutions are 1 mol/dm3.

I will include extra steps (in green) to the original solution, since its not clear.

Let volume of CH3COOH be x dm3
volume of CH3COONa = (0.1 - x) dm3

moles of CH3COOH = vol * concentration = x dm3 * 1 mol/dm3 = x mol
moles of CH3COONa = vol * concentration = (0.1 - x) dm3 * 1 mol/dm3 = (0.1 - x) mol

[CH3COONa] = moles of CH3COONa/volume of buffer = x/0.1
[CH3COOH] = moles of CH3COOH/volume of buffer = (0.1 - x)/0.1

[CH3COONa]/[CH3COOH]
= [x/0.1]/[(0.1-x)/0.1] (cancel the top and bottom 0.1)
= x/(0.1 -x)
 
Lola_sweet

Lola_sweet

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can someone please give me advice for organic chemistry?
i understand the concepts but its pastpapers tht kill me :/
any idea wht to do?
 
DeViL gURl B)

DeViL gURl B)

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T

Tasneem_m98

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Please help (M/J 2012 P11 number 22)

A sample of ethyl ethanoate is hydrolysed by heating under reflux with aqueous sodium hydroxide. The two organic products of the hydrolysis are separated, purified, and weighed.
Out of the total mass of products obtained, what is the percentage by mass of each product?

A) 32.4% and 67.6%
B) 38.3% and 61.7%
C) 42.3% and 57.7%
D) 50.0% and 50.0%

Thank you
 
Metanoia

Metanoia

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Tasneem_m98 said:
Please help (M/J 2012 P11 number 22)

A sample of ethyl ethanoate is hydrolysed by heating under reflux with aqueous sodium hydroxide. The two organic products of the hydrolysis are separated, purified, and weighed.
Out of the total mass of products obtained, what is the percentage by mass of each product?

A) 32.4% and 67.6%
B) 38.3% and 61.7%
C) 42.3% and 57.7%
D) 50.0% and 50.0%

Thank you
Click to expand...

You have misread the question, it should be ethyl propanoate.

Products are thus CH3CH2OH and CH3CH2COONa.

Using their Mr, % by mass of CH3CH2OH = 46/(46+96) =32.4%
 
cashew

cashew

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Lola_sweet said:
can someone please give me advice for organic chemistry?
i understand the concepts but its pastpapers tht kill me :/
any idea wht to do?
Click to expand...
just solve maximum pastpapers, that is the only possible good thing to do. Oh and make sure all of your concepts are clear
 
S

shazmina

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DeViL gURl B) said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w12_qp_42.pdf
question 3 part a (ii)
Click to expand...

Catalyst is a substance which speeds up the rate of a chemical reaction & is chemically unchanged at the end of the reaction
Example 1:- The Haber Process
N2 (g) + 3H2 (g) ------Fe (s) ------> 2NH3 (g)

Example 2:- S2O8^2- (aq) + 2I^1- (ap) ---------Fe^2+ ( Aq) --------> I2 (aq) + 2SO4^2-(aq)
The catalyst reduces the activation energy & provides an alternative pathway for the reaction to occur

REMEMBER THE ABOVE 2 EXAMPLES .. and thats all u have to write ..... no need to further explain about the state of the catalyst as it clearly shown in the balanced equation .........
 
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