Question

Calculate the pH at the equivalence point in titrating 0.090 M

solutions of each of the following with 0.075 M NaOH.
(a) perchloric acid (HClO4)
pH =
(b) boric acid (H3BO3), Ka =
5.8e-10
pH =

Answer 1

The pH at the equivalence point for titration of 0.090 M perchloric acid with 0.075 M NaOH is 7.00. For boric acid with a Ka of 5.8e-10, the pH will be slightly above 7 due to the hydrolysis of the conjugate base formed.

Step-by-step explanation:

To calculate the pH at the equivalence point in titrating 0.090 M solutions of different acids with 0.075 M NaOH, we should consider the nature of the acids involved.

(a) Perchloric Acid (HClO4)

Perchloric acid is a strong acid and will react completely with NaOH, a strong base, to produce water and a salt. At the equivalence point, the solution is neutral, and the pH is 7.00.

(b) Boric Acid (H3BO3)

Boric acid is a weak acid with a given Ka of 5.8e-10. To find the pH at the equivalence point when titrated with a strong base (NaOH), we need to recognize that the salt produced will hydrolyze to form OH− ions in water, slightly increasing the pH above 7. The concentration of OH− at the equivalence point can be calculated from the Kb of the conjugate base, and once we have [OH−], we can calculate the pOH and then the pH.

Answer 2

To calculate the pH at the equivalence point, for perchloric acid, it is 7.00 due to its strong acid nature. For boric acid with a Ka of 5.8e-10, the pH will be the pKa of the conjugate base, which is 9.24.

Step-by-step explanation:

Calculating pH at the Equivalence Point in Titration

To calculate the pH at the equivalence point in titrating 0.090 M solutions with 0.075 M NaOH:

• (a) For perchloric acid (HClO4), a strong acid, the solution at the equivalence point will be neutral, and thus the pH will be 7.00 because NaOH neutralizes the acid completely.
• (b) For boric acid (H3BO3), a weak acid with a given Ka of 5.8e-10, we must use the Henderson-Hasselbalch equation to find the pH. Since at the equivalence point, the acid has been fully converted to its conjugate base, the pH will be dictated by the hydrolysis of the conjugate base formed. The equation will be pH = pKa + log([A-]/[HA]), where pKa = -log(Ka), but in this case [A-] = [HA] because we are at the equivalence point, simplifying the equation to just pH = pKa.

In the case of boric acid, the pH will be equal to the pKa of the conjugate base formed during titration, thus:

• pKa = -log(5.8e-10) = 9.24
• pH = pKa = 9.24

Therefore, the pH at the equivalence point when titrating boric acid with NaOH will be 9.24.