Answer :
Final answer:
The pH at the second equivalence point in the titration of ascorbic acid with NaOH would be 8. This is because at the second equivalence point, all the ascorbic acid has been converted into its conjugate base, which hydrolyses in water to give a basic solution.
Explanation:
The second equivalence point in a double titration occurs when the second H+ from ascorbic acid is ionized and reacted completely with NaOH. In this case, the remaining species is the conjugate base of ascorbic acid, which is a weak base. As NaOH (a strong base) neutralizes both H+ ions of the ascorbic acid completely, we are left with a solution of water and salts. The pH is not 7, as one might guess, but rather depends on the relative strengths of the weak acid and the strong base.
At the second equivalence point, all ascorbic acid has been converted to its conjugate base, and the reaction can be represented as H2C6H6O6 - > 2H+ + C6H6O6^2-. Hence, the species that determines the pH of the solution is C6H6O6^2-, a weak base. As the weak base hydrolyses in water, this solution will be basic, and the pH at the second equivalence point will be greater than 7.
Therefore, among the given options, the pH at the second equivalence point in the titration of ascorbic acid with NaOH will be (C) 8.00. This is because the pH scale varies from 0 (acidic) to 14 (alkaline/basic) and for a weak acid-strong base titration, at the equivalence point, the pH is >7 (basic).
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