volumetric analysis

i need sum serious help with all these calculations.....i understand bits but am finding it hard to do back titrations, and sum other parts.....

like this question.......a solution of barium hydroxide contains 17.1g dm-3.
calculate its molarity........and the volume of this barium hydroxide solution which would be required to react exactly with 25cm3 of a solution containing 12g of ethanoic acid. CH3COOH per litre.
isnt 17.1g dm-3 the concentration? why does it wnt me 2 calculate the molarity then? if not, then do i simply do 17.1g dm-3 divide RMM?

here is a back titration example:

1.0g of limestone is allowed to react with 100cm3 of 0.2M HCL solution. when the reaction was complete the excess acid remaining required 24.8cm3 of 0.1M NaOH solution for neutralisation. calculate the percentage of calcium carbonate in the limestone.

could sum 1 plz explain how 2 do these step by step...am v.confused. thank u

Reply 1

Comp_Genius

4

urban_flavaz

i need sum serious help with all these calculations.....i understand bits but am finding it hard to do back titrations, and sum other parts.....

like this question.......a solution of barium hydroxide contains 17.1g dm-3.
calculate its molarity........and the volume of this barium hydroxide solution which would be required to react exactly with 25cm3 of a solution containing 12g of ethanoic acid. CH3COOH per litre.
isnt 17.1g dm-3 the concentration? why does it wnt me 2 calculate the molarity then? if not, then do i simply do 17.1g dm-3 divide RMM?

here is a back titration example:

1.0g of limestone is allowed to react with 100cm3 of 0.2M HCL solution. when the reaction was complete the excess acid remaining required 24.8cm3 of 0.1M NaOH solution for neutralisation. calculate the percentage of calcium carbonate in the limestone.

could sum 1 plz explain how 2 do these step by step...am v.confused. thank u

Molarity means no of moles per dm3 so you are right, divide by the RFM

Reply 2

mizfissy815

12

a) Molarity is the no. of moles in 1 dm3…they gave you the mass of barium hydroxide in 1 dm3….(both are considered concentration, but there values will be different).
Molar mass of Ba(OH)2 = 137+ (17*2)= 205

Moles= mass/molar mass
Moles= 17.1/205 = 0.0834
So the molarity is…0.0834mol dm-3.

b) Moles (Amount) = conc. * vol.
Vol.= moles/ conc.
Vol. = 0.0834/17.1 = 4.88*10^-3 dm3

~*~*~*~*~*~*~*~

2) Write the equations first…it helps
CaCO3 + 2HCl --> CaCl2 +H2O +CO2
HCl + NaOH --> NaCl + H2O

Molar Mass of CaCO3 = 40 +12 +(3*16) = 100
Moles (amount) of HCl unreacted (i.e. the excess) = Moles of NaOH used…as it is a 1 to 1 ratio.

Moles of NaOH= conc. * vol.
= 0.1 * 0.0248= 2.48*10^-3 moles
This is amount is equal to the no. moles in the excess/unreacted HCl. To find the moles that DID react…we have to find the amount in the original and then subtract the two.

Moles in Original HCL= conc. * vol.
= 0.2 * 0.1 = 0.02 moles
So the amount of moles of HCl that did react with the CaCO3.= original – unreacted
= 0.02 – 2.48*10^-3 = 0.01752 moles

Now that we found the amount of moles of HCl that reacted with CaCO3 , we can find the moles of the CaCO3, since the ratio of CaCO3 to HCl is 1:2
CaCO3 : HCl
1 : 2
x : 0.01752 Cross Multiply

= 0.01752/2 = 8.76*10^-3 moles of CaCO3 reacted.
Now we have to find the mass that reacted…
Mass= moles * molar mass
Mass= 8.76*10^-3 * 100 = 0.0876 grams

To find the percentage (purity) of CaCO3 in limestone =
(Mass of CaCO3 in sample / mass of the given sample)*100
= (0.876/1) *100 = 87.6%

:biggrin:

Plz feel free to correct any mistakes I might have made... :tongue:

Reply 3

supreme

4

fisfos815

a) Molarity is the no. of moles in 1 dm3…they gave you the mass of barium hydroxide in 1 dm3….(both are considered concentration, but there values will be different).
Molar mass of Ba(OH)2 = 137+ (17*2)= 205

Moles= mass/molar mass
Moles= 17.1/205 = 0.0834
So the molarity is…0.0834mol dm-3.

b) Moles (Amount) = conc. * vol.
Vol.= moles/ conc.
Vol. = 0.0834/17.1 = 4.88*10^-3 dm3

~*~*~*~*~*~*~*~

2) Write the equations first…it helps
CaCO3 + 2HCl --> CaCl2 +H2O +CO2
HCl + NaOH --> NaCl + H2O

Molar Mass of CaCO3 = 40 +12 +(3*16) = 100
Moles (amount) of HCl unreacted (i.e. the excess) = Moles of NaOH used…as it is a 1 to 1 ratio.

Moles of NaOH= conc. * vol.
= 0.1 * 0.0248= 2.48*10^-3 moles
This is amount is equal to the no. moles in the excess/unreacted HCl. To find the moles that DID react…we have to find the amount in the original and then subtract the two.

Moles in Original HCL= conc. * vol.
= 0.2 * 0.1 = 0.02 moles
So the amount of moles of HCl that did react with the CaCO3.= original – unreacted
= 0.02 – 2.48*10^-3 = 0.01752 moles

Now that we found the amount of moles of HCl that reacted with CaCO3 , we can find the moles of the CaCO3, since the ratio of CaCO3 to HCl is 1:2
CaCO3 : HCl
1 : 2
x : 0.01752 Cross Multiply

= 0.01752/2 = 8.76*10^-3 moles of CaCO3 reacted.
Now we have to find the mass that reacted…
Mass= moles * molar mass
Mass= 8.76*10^-3 * 100 = 0.0876 grams

To find the percentage (purity) of CaCO3 in limestone =
(Mass of CaCO3 in sample / mass of the given sample)*100
= (0.876/1) *100 = 87.6%

:biggrin:

Plz feel free to correct any mistakes I might have made... :tongue:

I get the same answer also.

Reply 4

supreme

4

urban_flavaz

i need sum serious help with all these calculations.....i understand bits but am finding it hard to do back titrations, and sum other parts.....

like this question.......a solution of barium hydroxide contains 17.1g dm-3.
calculate its molarity........and the volume of this barium hydroxide solution which would be required to react exactly with 25cm3 of a solution containing 12g of ethanoic acid. CH3COOH per litre.
isnt 17.1g dm-3 the concentration? why does it wnt me 2 calculate the molarity then? if not, then do i simply do 17.1g dm-3 divide RMM?

here is a back titration example:

1.0g of limestone is allowed to react with 100cm3 of 0.2M HCL solution. when the reaction was complete the excess acid remaining required 24.8cm3 of 0.1M NaOH solution for neutralisation. calculate the percentage of calcium carbonate in the limestone.

could sum 1 plz explain how 2 do these step by step...am v.confused. thank u

You should read a book called Calculations in AS/A Level Chemistry by Jim Clark.

The back titrations are quite easy really and I will try to explain it:

In the above question they said they had a known mass of limestone which they weighted but the problem is we don't know how much CaCO3 is in there do we? So the way to solve this is to do a titration. As we know how CaCO3 reacts with an acid. Think of it as though you are actually doing the experiment, you don't know accurately how much acid you need to add (as you don't know the purity of the limestone, the mass of CaCO3. Therefore you would use an indicator and keep adding the acid until it was in excess. The problem is now you don't know how much is in excess, so therefore you titrate the solution again against a standard solution of NaOH.

You know the number of moles of NaOH, as you added it, you know that HCl reacts with a 1:1 ratio. So the number of moles of NaOH, is the number of moles left over and because of the ratio mentioned above you also know the number of moles of HCl left over. Therefore you can calculate the number of moles of HCl which actually reacted by substracting the moles of excess HCl (you added this the first time) from the number of moles left over of HCl. Therefore you know the number of moles of HCl that reacted exactly. You know CaCO3 and HCl react in a 1:2 ratio so you half the number of moles of HCl that reacted then you calculate the mass of CaCO3 by multiplying the numbers of moles of CaC03 by its relative atomic mass. Then bingo! You have the mass of CaCO3 in the limestone, therefore you can calculate % mass by dividing the mass of CaCO3 by the mass of limestone and multiplying by a 100.

I hope this helps I used to find this hard at first.

PS: I apologise for any errors above.

volumetric analysis

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