Question

A sample of a certain monoprotic weak acid was dissolved in water and titrated with 16.00mL of 0.125M NaOH to reach the equivalence point. During the titration, the pH after adding 2.00mL of NaOH was 6.912. Calculate the Ka for the weak acid.

Answer 1

Ka = 8.572x10⁻⁷

Step-by-step explanation:

The equilibrium that takes places is:

• HA + OH⁻ ⇔ A⁻ + H₂O

At the equivalence point, all of the acid (HA) is converted into A⁻. Because the acid is monoprotic, the moles of OH⁻ are equal to the moles of the acid:

• 0.125 M * 16.00 mL = 2 mmol OH⁻ = 2 mmol HA

So the sample of the weak acid contains 2 mmol.

When 2.00 mL of the NaOH solution are added, some of the acid is turned into A⁻:

• 2.00 mL * 0.125 M = 0.25 mmol OH⁻

• So 0.25 mmol of A⁻ are formed, and (2 mmol - 0.25 mmol) 1.75 mmol of HA remain.

With the above information we can calculate the Ka, using the Henderson-Hasselbach equation:

• pH = pKa + log [(A⁻)/(HA)]

6.912 = pKa + log[0.25/1.75]

6.912 = pKa - 0.845

• pKa = 6.067

• Ka =
  `10^(-6.067)` = 8.572x10⁻⁷