June 01, 2020, 09:29:20 AM
Forum Rules: Read This Before Posting


Topic: Titrations  (Read 23106 times)

0 Members and 1 Guest are viewing this topic.

ssssss

  • Guest
Titrations
« on: October 14, 2004, 08:36:42 AM »
Ok i know basic about titration.In a sentence if we say:

Titration is used to find the concentration/streangth of the products in the mixture using laws of Equivalents.Indicators are nothing but the substances which gives readings by telling the extent of reactions.

Damn the titration is not as simple as it seems.
To have mastry on the numericals of titration the above matter is just not Enough.

1st of My problems is Equivalent weights of the Components?
Equivalent weights are Just not as simple.Like Equivalent weight of Na2CO3 is M/2 in presence of Methyl orange as Indicator whereas M when Phenolphatalein is used as Indicator[Because of Disproportionation reaction].

Can you Tell me What is the Value of Equivalent weight of Some common Strong Bases and Strong acids in Presence of Common Indicators like Methyl orange,Blue litmus,KMnO4 etc.

Also please tell which indicators are suitable For various titrations.Like strong acid and weak base,weak acid and strong base,strong acid and strong base,weak acid and weak base.

Lastly Please Explain this case of Double Titration

A mixture of Na2CO3 and NaHCO3 in 25ml.This mixture was titrated in presence of Phenolphatalein as indicator and With HCL acid of 0.1N.The reading came at 40ml.After this 1st reading 20ml of more acid of the same is put to get the second reading in presence of methyl orange as Indicator.What is the concentration of Both the Components?

Thanks.

Demotivator

  • Guest
Re:Titrations
« Reply #1 on: October 14, 2004, 09:07:03 AM »
Approximate values for ph ranges of indicators.

Phenolthalein- ph 8.5- 9.5
bromothymol blue - ph 6.2-8
methyl red ph 4.2-6
methyl orange ph 3-4

So depending on the acid base system, neutralization occurs at different phs.
Strong acid/strong base neutralize around ph 7. So, bromothymol blue. However, phenolthalein is also commonly used. (The slight offset in ph does not affect the result much because the change in ph is rapid for strong acid/base).
Same as above for weak acid/weak base

Weak acid/strong base : neutralization at pH > 7, therefore phenolthalein.

strong acid/weak base: neutralize at ph<7, methyl red or orange.

(I'll check the other problem, if i can, a bit later)

ssssss

  • Guest
Re:Titrations
« Reply #2 on: October 14, 2004, 09:12:51 AM »
Ok i got it.So i have to learn the Ph of some commonly used indicators.

Demotivator

  • Guest
Re:Titrations
« Reply #3 on: October 14, 2004, 09:24:49 AM »
There's something bugging me about the last problem. You sure 40 ml  and 20 ml for first and second titration? Doesn't make sense at first glance.
Na2CO3 -> NaHCO3
then the bicarbonate from the above along with the bicarbonate from the initial gets titrated. Seems to need more than 20 ml.
« Last Edit: October 14, 2004, 09:27:18 AM by Demotivator »

ssssss

  • Guest
Re:Titrations
« Reply #4 on: October 14, 2004, 09:28:10 AM »
There's something bugging me about the last problem. You sure 40 ml  and 20 ml for first titration? Doesn't make sense at first glance.
Na2CO3 -> NaHCO3
then the bicarbonate from the above along with the bicarbonate from the initial gets titrated. Seems to need more than 20 ml.

20 ml is for the second titration followed by the First.And currently i dont have the exact data from which i can Confirm.can you Just give the Equations for the 2 Titrations?

ssssss

  • Guest
Re:Titrations
« Reply #5 on: October 14, 2004, 09:31:23 AM »
Well i know inpresence of Phenolphatalein the Na2CO3 will react as follows.

Na2CO3 + HCL------->NaHCO3 + NaCl

Eq wt=M/1

Demotivator

  • Guest
Re:Titrations
« Reply #6 on: October 14, 2004, 09:35:17 AM »
The first titration converts Na2CO3 to NaHCO3:
Na2CO3 + H+ -> NaHCO3  phenolthalein ind.

The second one that goes to methyl orange is:
NaHCO3 + H+ -> H2CO3  + Na+

Do you see my problem?
If it takes 40 ml for the first titration, It should take at least 40 ml or more for the second titration because the same amount NaHCO3 has to be converted to H2CO3 as well as the NaHCO3 present in the original mixture.

ssssss

  • Guest
Re:Titrations
« Reply #7 on: October 14, 2004, 09:38:21 AM »
The first titration converts Na2CO3 to NaHCO3:
Na2CO3 + H+ -> NaHCO3  phenolthalein ind.

The second one that goes to methyl orange is:
NaHCO3 + H+ -> H2CO3  + Na+

Do you see my problem?
If it takes 40 ml for the first titration, It should take at least 40 ml or more for the second titration because the same amount NaHCO3 has to be converted to H2CO3 as well as the NaHCO3 present in the original mixture.



Well i want to say that the second Titration is followed by the First.So some of the Na2CO3 is already consumed.Isnt it?

Demotivator

  • Guest
Re:Titrations
« Reply #8 on: October 14, 2004, 09:42:37 AM »
Yeah, ALL the Na2CO3 has been consumed by the first. But it was consumed and turned to an equal amount of NaHCO3. So if it took 40 ml to consume Na2CO3, which is now NaHCO3, how can it take less than 40 ml to consume the NaHCO3 in the second titration?

ssssss

  • Guest
Re:Titrations
« Reply #9 on: October 14, 2004, 09:47:47 AM »
Yeah, ALL the Na2CO3 has been consumed by the first. But it was consumed and turned to an equal amount of NaHCO3. So if it took 40 ml to consume Na2CO3, which is now NaHCO3, how can it take less than 40 ml to consume the NaHCO3 in the second titration?

I got it.My book says that all of the of Na2CO3 is not consumed by the 1st titration.Instead all of the NaHCO3 is consumed leaving behind only left over Na2CO3 which will interact with HCl in second titration in presence Of methyl orange.

Demotivator

  • Guest
Re:Titrations
« Reply #10 on: October 14, 2004, 09:54:04 AM »
ok. I'll have to think about it later.

ssssss

  • Guest
Re:Titrations
« Reply #11 on: October 14, 2004, 09:55:47 AM »
Yeah.Its giving a migrane to me too.

Demotivator

  • Guest
Re:Titrations
« Reply #12 on: October 14, 2004, 10:28:55 PM »
1) Equiv wt does not depend on the indicator

2) equiv wt of Na2CO3 is always MW/2 because Na2CO3 has the capacity to accept two protons to fully neutralize (remember, eq wt is wt that can accept or give one mole (in this case H+) of something. Again, indicator like phenolthalein has nothing to do with it. Perhaps you're confusing a situation where the thing that's being titrated at a particular indicator is NaHCO3, in which case the eq wt of NaHCO3 is MW.

3) As for the last problem, I changed the original order of 40 ml then 20, to 20 ml then 40 because this makes more sense (unless someone can prove otherwise). The corrected problem:

A mixture of Na2CO3 and NaHCO3 in 25ml.This mixture was titrated in presence of Phenolphatalein as indicator and With HCL acid of 0.1N. The reading came at 20 ml. After this 1st reading 40 ml of more acid of the same is put to get the second reading in presence of methyl orange as Indicator. What is the concentration of Both the Components?

Na2CO3/NaHCO3 mixture in 25 ml:
Na2CO3 is the stronger base so it reacts first:
Na2CO3 + HCl -> NaHCO3 + NaCl  it takes 20 ml to do this (phenolth endpoint at PH about 9). Yet, notice, it has not completely reacted. The NaHCO3 that has just formed requires another 20 ml to form H2CO3. The total volume, therefore is 20+20 = 40 ml for complete reaction in the first part. This could not be deduced just from the titration end point volume if you did not know the nature of the chemical system you're working with. The calculation for this part is:

equivalents base = equivalents acid
equivalents Na2CO3 =(40/1000 L)(.1N) = .004 equiv
Moles Na2CO3 = (.004 equiv)(1 mole/2 equivalents) = .002 moles
Molarity Na2CO3 = .002 moles/.025 L = .08 M  

Now, 40 ml have been used up. That leaves 20 ml to titrate the original NaHCO3 in the mixture completely to H2CO3 at methyl orange.

equivalents base = equivalents acid
equivalents NaHCO3 =(20/1000 L)(.1N) = .002
Moles NaHCO3 = (.002 equiv)(1 mole/1 equivalent) = .002 moles
(notice that in this case it is 1 mole/1 equiv above)
Molarity NaHCO3 = .002 moles/.025 L = .08 M  
So the mixture has equal concentrayions of components.

« Last Edit: October 14, 2004, 10:30:20 PM by Demotivator »

ssssss

  • Guest
Re:Titrations
« Reply #13 on: October 15, 2004, 02:18:01 AM »
1) Equiv wt does not depend on the indicator

2) equiv wt of Na2CO3 is always MW/2 because Na2CO3 has the capacity to accept two protons to fully neutralize (remember, eq wt is wt that can accept or give one mole (in this case H+) of something. Again, indicator like phenolthalein has nothing to do with it. Perhaps you're confusing a situation where the thing that's being titrated at a particular indicator is NaHCO3, in which case the eq wt of NaHCO3 is MW.

3) As for the last problem, I changed the original order of 40 ml then 20, to 20 ml then 40 because this makes more sense (unless someone can prove otherwise). The corrected problem:

A mixture of Na2CO3 and NaHCO3 in 25ml.This mixture was titrated in presence of Phenolphatalein as indicator and With HCL acid of 0.1N. The reading came at 20 ml. After this 1st reading 40 ml of more acid of the same is put to get the second reading in presence of methyl orange as Indicator. What is the concentration of Both the Components?

Na2CO3/NaHCO3 mixture in 25 ml:
Na2CO3 is the stronger base so it reacts first:
Na2CO3 + HCl -> NaHCO3 + NaCl  it takes 20 ml to do this (phenolth endpoint at PH about 9). Yet, notice, it has not completely reacted. The NaHCO3 that has just formed requires another 20 ml to form H2CO3. The total volume, therefore is 20+20 = 40 ml for complete reaction in the first part. This could not be deduced just from the titration end point volume if you did not know the nature of the chemical system you're working with. The calculation for this part is:

equivalents base = equivalents acid
equivalents Na2CO3 =(40/1000 L)(.1N) = .004 equiv
Moles Na2CO3 = (.004 equiv)(1 mole/2 equivalents) = .002 moles
Molarity Na2CO3 = .002 moles/.025 L = .08 M  

Now, 40 ml have been used up. That leaves 20 ml to titrate the original NaHCO3 in the mixture completely to H2CO3 at methyl orange.

equivalents base = equivalents acid
equivalents NaHCO3 =(20/1000 L)(.1N) = .002
Moles NaHCO3 = (.002 equiv)(1 mole/1 equivalent) = .002 moles
(notice that in this case it is 1 mole/1 equiv above)
Molarity NaHCO3 = .002 moles/.025 L = .08 M  
So the mixture has equal concentrayions of components.




Well i truely dont know it is correct or not but it does Sound logical.These Indian books are so confusing.I tell you sir this book of mine says indicator can affect the Eq Wt. ohh....what a falicy?But anyway i am adapting your method and will solve some similar Titration numericals on this bases of this.I will let you know if i get any Contradictions anywhere.
I am going to buy i book on analytical chemistry of some reputed foreighn publications as early as possible.

Thanks.

Demotivator

  • Guest
Re:Titrations
« Reply #14 on: October 15, 2004, 12:43:39 PM »
Oh brother. This is really a matter of interpretation like is the glass half empty or half full? Answer: both.
I see now what they mean by the indicator affects the equiv wt. You can apply that logic and get the right answer.
I can also apply the logic that it does not affect and get the same answer because I compensate by understanding the extent of the reaction.
I'll give a simple example.
Just Na2CO3 in 25 ml solution titrated with .1N HCl. What is the moles Na2CO3?
The above can be titrated 2 different ways.

method 1: It takes 20 ml to titrate using phenolthalein.
method 2: It takes 40 ml to titrate using methyl orange.

In method 1, the endpoint is reached when all Na2Co3 turns to NaHCO3. (we end the analysis at this point).
In this case, their claim is Na2CO3 is 1 equiv/1 mole because it only consumes 1 H.
equiv Na2CO3 = (0.1N)(20/1000 L) = .002
moles Na2CO3 = 0.002 equiv(1 mole/1 equiv) = .002 moles
 
In method 2, the endpoint is reached when all Na2Co3 turns to H2CO3.
In this case, they would now claim that Na2CO3 is 2 equiv/1 mole because it consumes 2 H.
equiv Na2CO3 = (0.1N)(40/1000 L) = .004
moles Na2CO3 = .004 equiv(1 mole/2 equiv) = .002 moles

Obviously, the same result by two methods because the volumes and equiv wts were different. So you can say the indicator plays a role.
However, the way I was interpreting it was that the equiv wt was the same (MW/2) in both cases and  that in method 1, the reaction goes to HALF completion at the phenol endpoint, While in method 2 it goes to full completion. It's a different interpretation that leads to the same result.
 
For method 1 my calc:
equiv Na2CO3 = (0.1N)(2)(20/1000 L) = .004
Note the factor 2 above in recognition that the volume would be double for a complete reaction.
moles Na2CO3 = 0.004 equiv(1 mole/2 equiv) = .002 moles
(the same result as the other calc for method 1)

So When one simply asks, what is the equiv wt of Na2CO3, how to reply? If no further info is given, then apply the rules (for basic salts) and you get MW/2. This assumes a complete reaction using the full capacity of Na2CO3 as per definition of equivalent wt.
The problem is some people take a definition and "bend" it resulting in confusion.
« Last Edit: October 15, 2004, 06:20:47 PM by Demotivator »

Sponsored Links