The diprotic acid ascorbic acid has Ka values of 6.5 x 10-5 and 2.5 x 10-12. A 25.00mL 0.135M sample of ascorbic acid is titrated with 0.0987M NaOH. What is the pH of the solution after the addition of 55.0 mL of NaOH?

Answer 1

To solve this problem, first we need to find out which of the two Ka values will be used for the calculation. Since ascorbic acid is a diprotic acid, the first Ka will be used for the first dissociation, and the second for the second dissociation. In this case, the first dissociation is occurring, so we will use the first Ka value:

Ka1 = 6.5 x 10^-5

Next, we need to calculate the moles of ascorbic acid and moles of NaOH:

moles of ascorbic acid = 0.135 M * 0.025 L = 0.003375 moles
moles of NaOH = 0.0987 M * 0.055 L = 0.0054285 moles

Now, since NaOH is being added to the ascorbic acid, it will react with the hydrogen ions from the acid, and neutralize some portion of the acid:

moles of remaining ascorbic acid = moles of ascorbic acid - moles of NaOH
moles of remaining ascorbic acid = 0.003375 moles - 0.0054285 moles = -0.0020535 moles

As the moles of ascorbic acid is negative, this means all of the ascorbic acid has been neutralized and the solution has moved into the buffer region, where there is a mixture of the weak acid and its conjugate base.

In order to calculate the composition of the solution, we need to calculate the moles of ascorbic acid that have been converted into its conjugate base:

moles of conjugate base = moles of NaOH - moles of ascorbic acid reacted
moles of conjugate base = 0.0054285 moles - 0.003375 moles = 0.0020535 moles

The total volume of the solution now will be:

V_total = V_NaOH + V_ascorbic = 0.055 L + 0.025 L = 0.08 L

Now, we can calculate the concentrations of both ascorbic acid (HA) and its conjugate base (A-) in the solution:

[HA] = (0.003375 moles - 0.0020535 moles) / 0.08 L = 0.0165375 mol/L
[A-] = 0.0020535 moles / 0.08 L = 0.02566875 mol/L

Now, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution:

pH = pKa + log([A-]/[HA])
pH = -log(6.5 x 10^-5) + log(0.02566875/0.0165375)
pH = 4.19 + log(1.55246)
pH = 4.19 + 0.19075
pH = 4.38

Therefore, the pH of the solution after the addition of 55.0 mL of NaOH is 4.38.