What is the numerical relationship between the volume of base added to reach the first endpoint and the volume added to reach the final endpoint in the titration of a diprotic acid? What is the relationship for a triprotic acid?
If you are titrating with the same concn, then v1 = v2 = v3. Reasonably, if it takes a particular volume to remove the 1st H, then it must take twice that much to remove the second and three times that much to remove the third ond.
In the titration of a diprotic acid, the numerical relationship between the volume of base added to reach the first endpoint and the volume added to reach the final endpoint is generally a 1:2 ratio. This is because each mole of diprotic acid requires two moles of base for complete neutralization.
To understand this relationship, you need to know the stoichiometry of the chemical reaction between the acid and the base. Let's consider the generic diprotic acid H2A and a strong base NaOH. The balanced equation for the reaction can be written as follows:
H2A + 2NaOH → Na2A + 2H2O
From the balanced equation, you can see that each mole of diprotic acid (H2A) reacts with two moles of base (NaOH). Therefore, when reaching the first endpoint, where half of the acid has been neutralized, the volume of base added is half of the volume needed to reach the final endpoint.
Now, let's consider the titration of a triprotic acid, such as H3A, with the same base NaOH. The balanced equation for this reaction can be generalized as:
H3A + 3NaOH → Na3A + 3H2O
From the balanced equation, you can observe that each mole of triprotic acid (H3A) requires three moles of base (NaOH) for complete neutralization.
Therefore, for a triprotic acid, the numerical relationship between the volumes of base added to reach the first and final endpoints is generally a 1:3 ratio. This means that the volume of base needed to reach the first endpoint is one-third of the volume required to reach the final endpoint.