How Do You Determine the Equivalence Points in a Double Titration Problem?

[the-ever-kid]

Homework Statement

There was this Question in my book:

About 40 mls of 0.1 M solution of a compound sesqui Carbonate $(\mathrm{Na}_{2}\mathrm{CO}_3.\mathrm{NaHCO_3}.$$\mathrm{2H_2O})$ are titrated with $x$ mls of .1 M $\mathrm{HCl}$ in the presence of phenolphthalien.And the same compund was made to react with $y$ mls of the same solution of $\mathrm{HCl}$ in presence of methyl orange find $x$ & $y$

The attempt at a solution

Ok what i did was apply the fact that in the presense of phenolphthalien all of the $\mathrm{Na}_{2}\mathrm{CO}_3$ would react and only half of the $\mathrm{NaHCO_3}$ would would appear to have reacted.

the if 40mls of .1 M complex was given then it would have 8 milli equivalents of $\mathrm{Na}_{2}\mathrm{CO}_3$ and 4 milli equivalents of $\mathrm{NaHCO_3}$

as equivalents of $\mathrm{HCl}$ and the sum of equivalents $\mathrm{Na}_{2}\mathrm{CO}_3$ $\mathrm{NaHCO_3}$ is equal thus,

$8 + 2 = .1x$ thus x was 100mls

now for the second part i applied that all of the carbonates would show complete reaction in presence of methyl orange as it was acidic so :

$8+4=.1y$

thus y was 120 mls

My problem
my problem is that the second part is right but the first is wrong(it is 40 mls), when i asked my teacher she said that only half of sodium carbonate would react and none of the sodium bicarb would react ,how?

im comfused...

 

[Borek]

Write reactions taking place. What is the first step of the carbonate protonation?

 

[the-ever-kid]

carbonate protonation...hmmm... $\mathrm{CO_3^{-2}} + \mathrm{H^+} \rightleftharpoons \mathrm{HCO_3^-}$

 

[the-ever-kid]

and bicarb protonation is $\mathrm{HCO_3^-}~+~\mathrm{H^+}~\rightleftharpoons~\mathrm{H_2CO_3}$

 

[Borek]

OK.

When does the bicarbonate protonation start? Is it protonated together with carbonate, or after all carbonate got protonated to HCO₃^-?

 

[the-ever-kid]

after all the carbonate gets protonated...

 

[Borek]

Good. If so, is this:

in the presense of phenolphthalien all of the $\mathrm{Na}_{2}\mathrm{CO}_3$ would react and only half of the $\mathrm{NaHCO_3}$ would would appear to have reacted.

correct?

 

[the-ever-kid]

yes...

 

[Borek]

No, it is not. Phenolphthalein changes color after carbonate was protonated to hydrogencarbonate. Earlier presence of hydrogencarbonate doesn't matter at all.

 

[the-ever-kid]

but my teacher said that only half of the carbonate protonation would be detected.

 

[the-ever-kid]

and yes none of the bicarb ...*sorry for earlier on*

 

[Borek]

but my teacher said that only half of the carbonate protonation would be detected.

That's correct. "Half of the carbonate protonation" and "half of the bicarobonate protonation" are two completely different things. First makes sense (half of the carbonate protonation is protonation yielding hydrogencarbonate), second doesn't (you can't protonate hydrogencarbonate half way - that is, you can protonate half of the amount, but you will not get an equivalence point this way).

 

[the-ever-kid]

why is only half of the carbonate protonated why not a third or a fourth or two thirds?

 

[Borek]

No "half of the carbonate protonated" but "half of the carbonate protonation" - when all carbonate is converted to hydrogecarbonate. That's half of the possible protonation of carbonate, as its protonation is a two step process, and each stage ends at a very different pH (well separated equivalence points).

Hydrogencarbonate can be protonated only once, so there is only one equivalence point.

 

[the-ever-kid]

you mean to say that only half of the carbonate protonation is detected?

 

[Borek]

Yes.

Edit: see titration curve of the carbonate titrated with hydrochloric acid:

There are two distinct equivalence points, and with a correct choice of indicators you can detect each one separately.