Double titration and law of equivalence

[Krushnaraj Pandya]

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?

 

[Borek]

When methyl orange is used Na2CO3 is converted into NaCl + CO2 + H2O

No. After phenolphthaleine endpoint there were no Na₂CO₃ in the solution. CO₃^2- was converted into HCO₃^- and it is HCO₃^- that reacts in the second titration stage.

 

[Krushnaraj Pandya]

No. After phenolphthaleine endpoint there were no Na₂CO₃ in the solution. CO₃^2- 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

 

[Borek]

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.

 

[Krushnaraj Pandya]

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?

 

[Krushnaraj Pandya]

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?

 

[Borek]

Write equation for the reaction in which CO₃^2- is protonated to HCO₃^-.

 

[baldbrain]

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⁺ + SO₄^2- ......*
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

 

[Borek]

For this, 2(Eq. of NaOH)=Eq. of H₂SO₄

I suggest you check the definition of the equivalent.

 

[Krushnaraj Pandya]

Got it! Thanks a lot

 

[baldbrain]

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.
I was just saying from the point of view of the H⁺ & OH^- ions in the reaction.