When a diprotic acid, H2A, is titrated by NaOH, the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape.
(a) Notice that the plot has essentially two titration curves. If the first equivalence point occurs at 100.0 mL of NaOH added, what volume of NaOH added corresponds to the second equivalence point?
If it takes 100 mL to remove the first H ion it will take another 100 mL to remove the second one (for a total of 200 mL to remove both). So the first equivalence point is at 100.0 mL and the second equivalence point is at 200.0 mL.
Thank you!
In a titration of a diprotic acid, the protons are generally removed one at a time, resulting in a pH curve with two equivalence points. The first equivalence point occurs when the moles of NaOH added are equal to the moles of H2A present in the solution.
If the first equivalence point occurs at 100.0 mL of NaOH added, we can use this information to find the volume of NaOH added corresponding to the second equivalence point.
To determine the volume of NaOH added at the second equivalence point, we need to consider the stoichiometry of the reaction between the diprotic acid H2A and NaOH. Since H2A is diprotic, it can react with two moles of NaOH to form two moles of water and one mole of the sodium salt of H2A.
Therefore, at the first equivalence point, 1 mole of H2A reacts with 1 mole of NaOH. At the second equivalence point, all the remaining moles of H2A react with NaOH to form the sodium salt of H2A.
We know that the first equivalence point occurs at 100.0 mL of NaOH added. To find the volume of NaOH added at the second equivalence point, we need to determine the total moles of H2A present in the solution.
Since the stoichiometry of H2A and NaOH is 1:1, the moles of H2A at the first equivalence point would be equal to the moles of NaOH added at the first equivalence point.
To find this value, we need to know the concentration of NaOH used in the titration.
Once we have the moles of H2A at the first equivalence point, we can subtract this value from the total moles of H2A initially present in the solution to find the moles of H2A at the second equivalence point.
Finally, we can use the stoichiometry of H2A and NaOH (1:1) to find the volume of NaOH added at the second equivalence point.
To find the volume of NaOH added corresponding to the second equivalence point, we need to understand how titration curves work and how to identify equivalence points.
In a titration, a solution of known concentration (in this case, NaOH) is gradually added to a solution of unknown concentration (in this case, diprotic acid) until a reaction between the two is complete. The point at which the reaction is complete is called the equivalence point.
The pH curve for a diprotic acid titration generally has two equivalence points because diprotic acids have two acidic protons that can be removed. As NaOH is added to the diprotic acid, the protons are removed one at a time, resulting in two distinct curves on the pH plot.
The first equivalence point occurs when the first proton is completely neutralized (when all of the diprotic acid has reacted with NaOH). The volume of NaOH added at this point is given as 100.0 mL.
To determine the volume of NaOH added at the second equivalence point, we can use the fact that the total volume at the second equivalence point should be twice the volume at the first equivalence point. This is because each mole of diprotic acid requires two moles of NaOH for complete neutralization.
Therefore, to find the volume of NaOH added at the second equivalence point, we can multiply the volume at the first equivalence point by 2:
Volume at the second equivalence point = 100.0 mL * 2 = 200.0 mL.
So, the volume of NaOH added corresponding to the second equivalence point is 200.0 mL.