Polyprotic conjugates

[MedPR]

"For a diprotic acid such as H2CO3, the first proton removed corresponds to the second proton gained by the conjugate base."

Where does the first proton gained by the conjugate base come from? Water?

H2CO3 + H2O ----> H3O+ + HCO3-

HC3O- + H2O ----> H3O+ + CO32-

---

Members don't see this ad.

 

[Pisiform]

hmm I guess water:

HCO3-(aq) + H2O(l) -----> H2CO3(aq) + OH-(aq)

 

[MedPR]

Reading further in TBR gives this:

The correct relationships between pka and pkb for the two respective conjugate pairs (of H2CO3, or any diprotic acid) are pka1 + pkb2 = 14 and pka2 + pkb1 = 14.

Why isn't it pka1+pkb1 = 14?

If H2CO3 gives HCO3-, isn't HCO3- the conjugate base? It seems like TBR is saying that CO3 2- is the conjugate base of H2CO3.

 

[Pisiform]

I think HCO3- is still the conjugate base for H2CO3 but when calculating the pK for diprotic acid we go from Ka1 to Ka2 and Kb2 to Kb1. So we still have two conjugate pairs: H2CO3/HCO3- and HCO3-/CO32-. Look at these equation it would make much sense that way:

H2CO3 + H2O ----> HCO3- + H+ This is Ka 1

HCO3- + H2O -----> H+ + CO32- This is Ka 2

CO32- + H2O -----> HCO3- + OH- This is defined by Kb1

HCO3- + H2O ------> H2CO3 + OH- This is defined by Kb2

so you have to go step wise from the last conjugate base produced to the first.

Thats why Kw = Ka1 * Kb2 & also Kw = Ka2 * Kb1

ans so the same goes for pka

 

[IH8ColdWeath3r]

yeah this is correct and this is from TBR.

Pka1 + Pkb 2 = 14
Pka2 + Pkb1 = 14

for diprotic acids. This is just saying that the Pka of the first proton plus the Pkb of gaining the second proton should equal to 14, and this hold true based on the equations above. This is when H2CO3 goes to HCO3- (its conjugate base) and HCO3- gains its SECOND PROTON (PKb2) to become H2CO3.

This holds true for triprotic acids like phosphoric acid

Pka1 +Pkb3 = 14
Pka2 + Pkb2 = 14
Pka3 + Pkb3 = 14

I wouldn't worry too much about this though.

 

[MedPR]

I think HCO3- is still the conjugate base for H2CO3 but when calculating the pK for diprotic acid we go from Ka1 to Ka2 and Kb2 to Kb1. So we still have two conjugate pairs: H2CO3/HCO3- and HCO3-/CO32-. Look at these equation it would make much sense that way:

H2CO3 + H2O ----> HCO3- + H+ This is Ka 1

HCO3- + H2O -----> H+ + CO32- This is Ka 2

CO32- + H2O -----> HCO3- + OH- This is defined by Kb1

HCO3- + H2O ------> H2CO3 + OH- This is defined by Kb2

so you have to go step wise from the last conjugate base produced to the first.

Thats why Kw = Ka1 * Kb2 & also Kw = Ka2 * Kb1

ans so the same goes for pka

Thanks, that makes sense.

yeah this is correct and this is from TBR.

Pka1 + Pkb 2 = 14
Pka2 + Pkb1 = 14

for diprotic acids. This is just saying that the Pka of the first proton plus the Pkb of gaining the second proton should equal to 14, and this hold true based on the equations above. This is when H2CO3 goes to HCO3- (its conjugate base) and HCO3- gains its SECOND PROTON (PKb2) to become H2CO3.

This holds true for triprotic acids like phosphoric acid

Pka1 +Pkb3 = 14
Pka2 + Pkb2 = 14
Pka3 + Pkb3 = 14

I wouldn't worry too much about this though.

Yea that's where I got confused today. I was reading that chapter in TBR and I didn't understand why the first proton lost from the acid is the second proton gained by the conjugate base.

TBR didn't really explain it in a way that I could comprehend.