Back-titration question
The question is
1.435g sample of dry CaCO^3 and CaCl^2 mixture was dissolved in 25.00mL of 0.9892M HCl solution. What was the CaCl^2 percentage in the original sample if 21.48dm^3 of 0.09312M of NaOH was used to titrate excess HCl??
We haven't been taught this thoroughly enough to understand and it would be a big help if someone could go through the working aswell as the answer!
Thanks!
1.435g sample of dry CaCO^3 and CaCl^2 mixture was dissolved in 25.00mL of 0.9892M HCl solution. What was the CaCl^2 percentage in the original sample if 21.48dm^3 of 0.09312M of NaOH was used to titrate excess HCl??
We haven't been taught this thoroughly enough to understand and it would be a big help if someone could go through the working aswell as the answer!
Thanks!
The two important reactions here are:
1) NaOH + HCl --> H2O + NaCl (you should know this one)
2) CaCO3 + 2HCl --> CaCl2 + CO2 + H2O (this one is less obvious)
The important thing to see here is that the CaCO3 will react, whereas the CaCl2 is not involved in any reactions at all.
As the name "back titration" implies, it's best to work backwards from the last bits of information given to get to the answer.
The first thing to do is work out the amount (mol) of NaOH that was used in the final step. You can use n=cv (remember to convert cm3 to dm3 by dividing by 1000)
Using reaction 1, you know that this must be the amount of HCl that was left in the solution, as NaOH and HCl are in a 1:1 ratio in reaction 1.
Now you should work out the amount of HCl in the original solution, using n=cv.
You know the original amount and how much was left after the sample was added, so you can work out the amount that was used to react with the sample.
From reaction 2, we can see that whatever the HCl amount was, only half as much CaCO3 reacted (they have a 1:2 ratio). This tells you the amount of CaCO3 in the sample.
You'll need the molar mass of CaCO3 - you should be able to easily get this from the periodic table. Now you can work out the mass of the CaCO3 amount that we calculated, which is the mass of it in the original sample. Now you just need to calculate the proportion of the sample that was CaCO3, and subtract this from 1 to get the proportion of CaCl2 since they were the only chemicals in the sample.
Answers:
1) NaOH + HCl --> H2O + NaCl (you should know this one)
2) CaCO3 + 2HCl --> CaCl2 + CO2 + H2O (this one is less obvious)
The important thing to see here is that the CaCO3 will react, whereas the CaCl2 is not involved in any reactions at all.
As the name "back titration" implies, it's best to work backwards from the last bits of information given to get to the answer.
The first thing to do is work out the amount (mol) of NaOH that was used in the final step. You can use n=cv (remember to convert cm3 to dm3 by dividing by 1000)
Using reaction 1, you know that this must be the amount of HCl that was left in the solution, as NaOH and HCl are in a 1:1 ratio in reaction 1.
Now you should work out the amount of HCl in the original solution, using n=cv.
You know the original amount and how much was left after the sample was added, so you can work out the amount that was used to react with the sample.
From reaction 2, we can see that whatever the HCl amount was, only half as much CaCO3 reacted (they have a 1:2 ratio). This tells you the amount of CaCO3 in the sample.
You'll need the molar mass of CaCO3 - you should be able to easily get this from the periodic table. Now you can work out the mass of the CaCO3 amount that we calculated, which is the mass of it in the original sample. Now you just need to calculate the proportion of the sample that was CaCO3, and subtract this from 1 to get the proportion of CaCl2 since they were the only chemicals in the sample.
Answers:
Spoiler
Please recheck the units in the question, I suppose they are wrong. The moles of HCl in the original reaction are (25/1000) x 0.9892 which is 0.02473 moles. However, if you calculate moles of NaOH using 21.48 dm^3 you are bound to get 2 which is higher than the moles of HCl present in the first place.Since the ration of NaOH : HCl is 1:1, it is impossible to be reacting EXCESS HCl with 2 moles of NaOH.
Original post by snehalb
Please recheck the units in the question, I suppose they are wrong. The moles of HCl in the original reaction are (25/1000) x 0.9892 which is 0.02473 moles. However, if you calculate moles of NaOH using 21.48 dm^3 you are bound to get 2 which is higher than the moles of HCl present in the first place.Since the ration of NaOH : HCl is 1:1, it is impossible to be reacting EXCESS HCl with 2 moles of NaOH.
I don't know what you have done but I managed to get a reasonable answer. This is what I did:
CaCO3 + 2HCL -> Cacl2 + CO2 + H2O
HCL + NaOH -> Nacl + H2O
moles of NaOH used in titration: 21.48/1000 x 0.09312 = 2.0002176 x 10^-3 mol
moles of HCl in titration = 2.0002176 x 10^-3
moles of HCL initially = 25/1000 x 0.9892
moles of HCL that reacted = (25/1000x 0.9892) -(2.0002176 x 10^-2) = 0.0227297824 mol
moles of CaCO3 = 0.227297824/2 = 0.0113648912 mol
mass of CaCO3 = 0.0113648912 x 100 (Mr of CaCO3) = 1.13648912
percentage by mass of CaCO3 = 1.13648912/1.435 x 100 = 79.20% (2dp)
Original post by theo_.pinto
I don't know what you have done but I managed to get a reasonable answer. This is what I did:
CaCO3 + 2HCL -> Cacl2 + CO2 + H2O
HCL + NaOH -> Nacl + H2O
moles of NaOH used in titration: 21.48/1000 x 0.09312 = 2.0002176 x 10^-3 mol
moles of HCl in titration = 2.0002176 x 10^-3
moles of HCL initially = 25/1000 x 0.9892
moles of HCL that reacted = (25/1000x 0.9892) -(2.0002176 x 10^-2) = 0.0227297824 mol
moles of CaCO3 = 0.227297824/2 = 0.0113648912 mol
mass of CaCO3 = 0.0113648912 x 100 (Mr of CaCO3) = 1.13648912
percentage by mass of CaCO3 = 1.13648912/1.435 x 100 = 79.20% (2dp)
CaCO3 + 2HCL -> Cacl2 + CO2 + H2O
HCL + NaOH -> Nacl + H2O
moles of NaOH used in titration: 21.48/1000 x 0.09312 = 2.0002176 x 10^-3 mol
moles of HCl in titration = 2.0002176 x 10^-3
moles of HCL initially = 25/1000 x 0.9892
moles of HCL that reacted = (25/1000x 0.9892) -(2.0002176 x 10^-2) = 0.0227297824 mol
moles of CaCO3 = 0.227297824/2 = 0.0113648912 mol
mass of CaCO3 = 0.0113648912 x 100 (Mr of CaCO3) = 1.13648912
percentage by mass of CaCO3 = 1.13648912/1.435 x 100 = 79.20% (2dp)
Find percentage of mass of cacl2 so it is 20.7 percent
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