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Topic: Titrations  (Read 23114 times)

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ssssss

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Re:Titrations
« Reply #15 on: October 16, 2004, 03:00:02 AM »
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.



Thats what i was talking.

In phenolphatalein as indicator:

Na2CO3 + HCL=>NaHCO3 + NaCl[careful,Na2CO3 is not competely changed to H2CO3 or H2O+CO2]

In methyl orange as indicator:

Na2CO3 + 2HCL=>2NaCL + H20 + CO2[Here Na2CO3 is complete changed to H20 and CO2
Note:H2CO3 being unstable will change to H2O + CO2,Also you can clearly see that here are 2 equivalents of HCL with one Na2CO3 which is Not as in the 1st case.


Now tell me that is Phenolphatalein will be suitable for the Titration in NaHCO3 Or not.

Now the real Problem about Double Titration:

25 ml of Na2CO3 and NaOH is titrated with Phenolphatalein as indicator and using 0.1N HCL.Reading came at 40ml.After this reading,to obtain the second reading an Indicator methyl orange is used and it was Found that 15ml of the same acid is used.What is the steangth of Na2CO3 and NaOH in the solution?


Ok now we can see that Na2CO3 and NaOH are both suitable For the Titration with Phenolphatalein.

So according to the method the 1st equation is:
Suppose there are a gms of NaOH and b gms of Na2CO3.

a/40 + b/106 = 4/1000 Eq.

Now for the second Equation we 1st Observe that the NaOH has given 1st reading and is comletelely consumed and Observe that the Na2CO3 is Completely Converted to NaHCO3.Now as we are taking second reading that means we are only Dealing with this NaHCO3.

Now also Observe that the NaHCO3 in the formed in the 1st reaction is b/106 Eq.
So this NaHCO3 takes 1.5Meq.

Therefore b/106=1.5/1000

Solving the 2 the result is:
b=0.159 gms/25ml
a=0.65 gms/25ml

ssssss

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Re:Titrations
« Reply #16 on: October 16, 2004, 03:01:16 AM »
I have surpassed the Titrations Now.

Demotivator

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Re:Titrations
« Reply #17 on: October 17, 2004, 06:14:02 PM »
Very good.
But the problem you have just stated:
25 ml of Na2CO3 and NaOH etc

Is not what you originally stated:
A mixture of Na2CO3 and NaHCO3 in 25ml etc
etc.

As I suspected, something was wrong with the problem.
« Last Edit: October 17, 2004, 06:40:47 PM by Demotivator »

Offline AWK

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Re:Titrations
« Reply #18 on: October 18, 2004, 02:24:13 AM »
The data are wrong. For pure Na2CO3 you need 40 (the first titration) + 40 (the second titration) mL of NaOH solution .
For pure NaHCO3 you need 0 ml in the first titration.
May be the second number should be 120 instead of 20, then mixture of Na2CO3 and NaHCO3 can be equimolar.
« Last Edit: October 18, 2004, 02:25:15 AM by AWK »
AWK

ssssss

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Re:Titrations
« Reply #19 on: October 18, 2004, 05:49:02 AM »
The data are wrong. For pure Na2CO3 you need 40 (the first titration) + 40 (the second titration) mL of NaOH solution .
For pure NaHCO3 you need 0 ml in the first titration.
May be the second number should be 120 instead of 20, then mixture of Na2CO3 and NaHCO3 can be equimolar.


Which problem are you talking about,one i gave before or one i have solved above?

ssssss

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Re:Titrations
« Reply #20 on: October 18, 2004, 05:52:37 AM »
Very good.
But the problem you have just stated:
25 ml of Na2CO3 and NaOH etc

Is not what you originally stated:
A mixture of Na2CO3 and NaHCO3 in 25ml etc
etc.

As I suspected, something was wrong with the problem.


Yeah i got it Demotivator.You were right.In case of Phenolphatalein,NaHCO3 is not taken into account,and so the second reading which involved the methyl orange should be more than the 1st reading.

Offline AWK

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Re:Titrations
« Reply #21 on: October 18, 2004, 08:19:27 AM »
I talked about the first (subject) problem posted by demotivator
AWK

faisaltheonly1

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Re:Titrations
« Reply #22 on: March 21, 2006, 10:55:02 AM »
hello,

iam interested on this topic as iam going to do that experiment, too.
(i am talking about NaOH & Na2CO3.)
but i got a question.
if
V1 = volume of HCL used in 1st titration with phenolphthalein
V2 = volume of HCL used in 1st titration with methyl orange.
V3 = V1- V2
then can; ???
V3 be used to find the concentration of NaOH? and
can 2 * V3 be used to find the concentration of Na2CO3?

Faisal

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