FFR: Polyprotic Titrations

Overall stoichiometric neutralization: H₂CO₃+2 OH⁻→CO₃²⁻+2 H₂O. Sodium ions are spectators if NaOH supplies the base.

The titration shown here, of $H_2CO_3$, has two equivalence points, as there are two neutralization reactions. The overall neutralization reaction is:

First step: H₂CO₃+OH⁻→HCO₃⁻+H₂O. One mole of hydroxide per mole of carbonic acid reaches the first stoichiometric equivalence point.

However, this reaction occurs in two steps:

1. $H_2CO_3+NaOH\leftrightarrow HCO_3^-+H_2O+Na^+$
2. $HCO_3^-+NaOH\leftrightarrow CO_3^{2-}+Na^++H_2O$

The titration equation we are used to has to be modified slightly to account for the double-deprotonation going on here.

In the first reaction, we can see that the normal situation is occurring. All the $NaOH$ we have added has reacted completely with the $H_2CO_3$, turning it completely into $HCO_3^-$. Therefore, for that point, we have only removed one proton:

At each equivalence point the added titrant has consumed a specified number of acid equivalents. Whether separate pH-curve jumps are resolved depends on the separated acid strengths, concentrations and measurement conditions. Counting protons alone does not guarantee visible jumps.

$n_{NaOH,\ added}=n_{H_2CO_3,\ initial}$

$C_{NaOH}V_{NaOH,\ added}=C_{H_2CO_3,\ initial}V_{H_2CO_3,\ initial}$

Second step: HCO₃⁻+OH⁻→CO₃²⁻+H₂O. The complete ionic form would place any Na⁺ spectator on both sides.

Combining the two steps gives H₂CO₃+2 OH⁻→CO₃²⁻+2 H₂O. Both atom totals and charge are conserved. The corresponding formal dominant acid-family form changes as base equivalents are added.

This reaction is not a $1:1$ ratio for the acid and base, therefore the equivalence formula is a bit different:

$n_{NaOH,\ added}=2\cdot n_{H_2CO_3,\ initial}$

$C_{NaOH}V_{NaOH,\ added}=2\cdot C_{H_2CO_3,\ initial}V_{H_2CO_3,\ initial}$

At the second equivalence point the formal product is carbonate. Subsequent equilibrium with water still produces some bicarbonate and hydroxide; “complete stoichiometric neutralization” does not mean every other acid-family form is literally absent.

Some examples of equivalence formulas for titrations:

Titration of $H_3PO_4$ with $NaOH$:

Equivalence Point 1:
$C_{NaOH}V_{NaOH,\ added}=C_{H_3PO_4,\ initial}V_{H_3PO_4,\ initial}$

Equivalence Point 2:
$C_{NaOH}V_{NaOH,\ added}=2\cdot C_{H_3PO_4,\ initial}V_{H_3PO_4,\ initial}$

Equivalence Point 3:
$C_{NaOH}V_{NaOH,\ added}=3\cdot C_{H_3PO_4,\ initial}V_{H_3PO_4,\ initial}$

Titration of $SO_3^{2-}$ with $HBr$:

Equivalence Point 1:
$C_{HBr}V_{HBr,\ added}=C_{SO_3^{2-},\ initial}V_{SO_3^{2-},\ initial}$

Equivalence Point 2:
$C_{HBr}V_{HBr,\ added}=2\cdot C_{SO_3^{2-},\ initial}V_{SO_3^{2-},\ initial}$

Chapter Problems

1. Problem 1