Question

Consider the titration of 0.100 L of 0.200 M ascorbic acid with 0.500 M NaOH. What is the pH at the endpoint of the titration? Enter your answer to two decimal places; for this question we can assume the 5% assumption is valid.

Answer 1

This is a case of titration of a weak acid with a strong base. Let us represent the reaction going on here as

HA + OH^- --------> A^- + H₂O

Let us find out the amount of ascorbic acid present in the solution in mole.

Amount of ascorbic acid = (0.100 L). (0.200 mole/L) = 0.02 mole

NaOH reacts with ascorbic acid on a 1:1 molar ratio; so 0.02 mole of ascorbic acid will be neutralized by 0.02 mole of NaOH.

The volume of NaOH required is (0.02 mole)/(0.500 mole/L) = 0.04 L

At the end point, we have sodium ascorbate and water. We calculate the concentration of sodium ascorbate as (0.02 mole)/(0.14 L) = 0.1428 M (ascorbate is obtained from ascorbic acid on a 1:1 molar ratio).

Now, ascorbate is the conjugate base of a weak acid and will undergo the following equilibrium reaction.

A^- + H₂O ----------> HA + OH^-

Initial                           0.1428                          0       0

Change                         - x                                 x       x

Equilibrium             (0.1428 – x)                   (0+x)(0+x)

The equilibrium constant

K_b = (x).(x)/(0.1428 – x)

Now, the acid dissociation constant for ascorbic acid is K_a = 7.9 x 10^-5

We also know K_a.K_b = K_w = 10^-14

Hence, K_b = K_w/K_a = (1.0 x 10^-14)/(7.9 x 10^-5) = 1.2658 x 10^-10

Hence, K_b = x²/(0.1428) (We assume x to be very small)

or, x² = K_b.(0.1428) = (1.2658 x 10^-10).(0.1428) = 1.8075 x 10^-11

or, x =4.2515 x 10^-6

So, the concentration of hydroxide in the solution at the end point is 4.2515 x 10^-6.

Now, pOH = - log₁₀[OH^-] = -log₁₀(4.2515 x 10^-6) = 5.37

Therefore, pH = 14 – pOH = 14 – 5.37 = 8.63

Ans: The pH at the end point is 8.63.