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Double titration and law of equivalence

  • #1
Krushnaraj Pandya
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Homework Statement


1) for a given reaction to consume one reactant completely, must the equivalents of both reactants be same? for example, I know in the reaction of HCl + NaOH - the equivalents of HCl=equivalents of NaOH for a titration, is it the same for Na2CO3 + HCl?
2) the following is an excerpt from my textbook-
Double titration: This is a titration of specific compound using different indicators. When the solution containing NaOH and Na2CO3 is titrated using phenolpthalein indicator, the following reaction takes place at the phenolpthalein end point.
NaOH + HCl - NaCl + H2O
Na2CO3 + HCl - NaHCO3 + NaCl
Here, Equivalents of NaOH + 1/2 equivalents of Na2CO3= equivalents of HCl....(i)
When methyl orange is used Na2CO3 is converted into NaCl + CO2 + H2O
Hence equivalents of NaOH + eqs. of Na2CO3=eqs. of HCl...(ii)

I dont't understand how we got the first equation and this titration in general, can someone please explain?
 
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Answers and Replies

  • #2
Borek
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When methyl orange is used Na2CO3 is converted into NaCl + CO2 + H2O
No. After phenolphthaleine endpoint there were no Na2CO3 in the solution. CO32- was converted into HCO3- and it is HCO3- that reacts in the second titration stage.
 
  • #3
Krushnaraj Pandya
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No. After phenolphthaleine endpoint there were no Na2CO3 in the solution. CO32- was converted into HCO3- and it is HCO3- that reacts in the second titration stage.
ok...but my doubts still remain,can you explain how we got both these equations and the titration in some detail? I'd be really grateful, thank you
 
  • #4
Borek
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Not clear to me what it is that you don't understand, it is just about knowing what reacts and balancing reaction equations.

What reacts can be predicted knowing the final pH and Ka values of acids present.
 
  • #5
Krushnaraj Pandya
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Not clear to me what it is that you don't understand, it is just about knowing what reacts and balancing reaction equations.

What reacts can be predicted knowing the final pH and Ka values of acids present.
why is the coefficient of Na2CO3 in the first equation 1/2? I know its n-factor is 2 but how does that come into play here?
 
  • #6
Krushnaraj Pandya
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why is the coefficient of Na2CO3 in the first equation 1/2? I know its n-factor is 2 but how does that come into play here?
hello?
 
  • #7
Borek
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Write equation for the reaction in which CO32- is protonated to HCO3-.
 
  • #8
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for a given reaction to consume one reactant completely, must the equivalents of both reactants be same?For example, I know in the reaction of HCl + NaOH - the equivalents of HCl=equivalents of NaOH
No.
H2SO4 +2NaOH →Na2SO4 + 2H2O
For this, 2(Eq. of NaOH)=Eq. of H2SO4
This is because H2SO4 ionizes as
H2SO4→2H+ + SO42- ......................*
Hence, as you see, for each mole of H2SO4, there are 2 moles of H+ ions formed. So, you need an equal quantity (2 mol) of OH- ions (to form the water), which are furnished by NaOH. And hence,
2(Eq. of NaOH)=Eq. of H2SO4
*You will understand this better after you do the topic of Chemical & Ionic Equilibria
 
  • #9
Borek
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For this, 2(Eq. of NaOH)=Eq. of H2SO4
I suggest you check the definition of the equivalent.
 
  • #10
Krushnaraj Pandya
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Got it! Thanks a lot
 
  • #11
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I suggest you check the definition of the equivalent.
All right, my bad. Let's just say that NaOH & H2SO4 react in the ratio 2:1.:oldsmile:
I was just saying from the point of view of the H+ & OH- ions in the reaction.
 
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