grahammnmr
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#1
Report Thread starter 6 years ago
#1
Hi,

Quite frankly, I'm stuggling with these titration calculations. Any ideas?

Attempted

Calculate the amount in mol of HCl in the volumetric flask

3.50/250 x 2 = 0.028mol

Calculate amount in mol of HCl used

0.028 x 36.2 (average titre) = 1.0136x10-3 mol
1000


No Ideas

Calculate the amount, in mol, of Ca(OH)2 reacting.

Amount in mol of Ca(OH)2 in 1dm3

Mass of Ca(OH)2 in 1dm3
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Malgorithm
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#2
Report 6 years ago
#2
(Original post by grahammnmr)
Hi,

Quite frankly, I'm stuggling with these titration calculations. Any ideas?

Attempted

Calculate the amount in mol of HCl in the volumetric flask

3.50/250 x 2 = 0.028mol

Calculate amount in mol of HCl used

0.028 x 36.2 (average titre) = 1.0136x10-3 mol
1000


No Ideas

Calculate the amount, in mol, of Ca(OH)2 reacting.

Amount in mol of Ca(OH)2 in 1dm3

Mass of Ca(OH)2 in 1dm3
Could you be a bit clearer with what data you obtained from the experiment?
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Dylann
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#3
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(Original post by grahammnmr)
Hi,

Quite frankly, I'm stuggling with these titration calculations. Any ideas?

Attempted

Calculate the amount in mol of HCl in the volumetric flask

3.50/250 x 2 = 0.028mol

Calculate amount in mol of HCl used

0.028 x 36.2 (average titre) = 1.0136x10-3 mol
1000


No Ideas

Calculate the amount, in mol, of Ca(OH)2 reacting.

Amount in mol of Ca(OH)2 in 1dm3

Mass of Ca(OH)2 in 1dm3
This is pretty unclear, could you post the exact questions and all details of the experiment. Your first section of working out doesn't make sense, moles x vol giving moles again?

(Original post by Malgorithm)
Could you be a bit clearer with what data you obtained from the experiment?
From what I can make of it, he's neutralising Ca(OH)2 by adding HCl, or the other way round, can't work out from the wording. Either way:

2HCl + Ca(OH)2 --> CaCl2 + 2H2O

Other than that, I don't agree with his mole calculations (mol = conc x vol, he divided by 250 for some reason?)
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grahammnmr
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#4
Report Thread starter 6 years ago
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Okay Doky,

So...

We're finding the conc of limewater by titration

HCl - 2mol ( i used 3.5cm3)

3.50 cm3 of HCl is diluted in a volumetric flask (250cm3) with water
I pippeted 25cm3 of limewater into a conical flask

For the 'Calculate the amount in mol of HCl in the volumetric flask'

i used n=cv
C= 2.00
V = 3.5/250 (I calculated the dilution)
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Dylann
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Report 6 years ago
#5
(Original post by grahammnmr)
Okay Doky,

So...

We're finding the conc of limewater by titration

HCl - 2mol ( i used 3.5cm3)

3.50 cm3 of HCl is diluted in a volumetric flask (250cm3) with water
I pippeted 25cm3 of limewater into a conical flask

For the 'Calculate the amount in mol of HCl in the volumetric flask'

i used n=cv
C= 2.00
V = 3.5/250 (I calculated the dilution)
Okay this makes a bit more sense!

So you're told to work out the amount of moles of HCl in the volumetric flask...well the amount of moles of HCl in the flask won't change from the original 3.5cm3 (it is just less concentrated in water)

moles = concentration x vol
= 2 x 3.5/1000

remember when we're calculating, we use volumes in dm3. To convert from cm3 to dm3 you divide by 1000. E.g. 3.5cm3 = 3.5/1000 dm3 or 0.0035dm3

so we have 0.007 moles of HCl

The "dilution" doesn't affect the amount of moles of HCl

I think you seem to have calculated the concentration of HCl in the flask, that would be:

conc = mol/vol
conc = 0.007/(253/1000)
=0.0276

remember your total volume is 253.5 not 250 as you have added 3.5cm3 of HCl

I think from your original post you said your avg titre was 36.5cm

Since half the moles of Ca(OH)2 react than that of HCl, 0.0035 moles of Ca(OH)2 react.

This means there were 0.0035 moles in the 36.5cm3 of Ca(OH)2, so the concentration was about 0.096mol-1dm3. Thus you'd expect 0.096 mol in 1dm3 of that Ca(OH)2 mixture..

mass = mr x mol
mr = 40.1 + 32 + 2 = 74.1
mol = 0.096

mr x mol = 74.1 x 0.096

= 7.11g
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