DETERMINING THE K_A OF AN UNKNOWN ACID

Before we do that, I’m going to derive the Henderson-Hasselbalch Equation (a very useful form of the K_a equation)

2

Look carefully at the following titration curve. We are going to use it to identify the substance being titrated.

3

What type of substance is the analyte

4

What type of substance is the titrant?

5

At 0 mL of NaOH added, what is present in the reaction flask?

6

HA + OH^- 🡪

HOH + A^-

At 0 mL of NaOH added, what is present in the reaction flask?

7

HA + OH^- 🡪

HOH + A^-

HA

only

At 5 mL of NaOH added, what substances are present in the reaction flask?

8

HA + OH^- 🡪

HOH + A^-

HA

only

At 5 mL of NaOH added, what substances are present in the reaction flask?

9

HA + OH^- 🡪

HOH + A^-

A lot of HA and a little A^-

HA

only

At 20 mL of NaOH added, what substances are present in the reaction flask?

10

HA + OH^- 🡪

HOH + A^-

A lot of HA and a little A^-

HA

only

At 20 mL of NaOH added, what substances are present in the reaction flask?

11

HA + OH^- 🡪

HOH + A^-

A lot of HA and a little A^-

A lot of A^- and a little HA

HA

only

At 25 mL of NaOH added, what substances are present in the reaction flask?

12

HA + OH^- 🡪

HOH + A^-

A lot of HA and a little A^-

A lot of A^- and a little HA

HA

only

At 12.5 mL of NaOH added (the half equivalence point), what substances are present in the reaction flask?

13

HA

only

HA + OH^- 🡪

HOH + A^-

A lot of HA and a little A^-

A lot of A^- and a little HA

A^- only

At 12.5 mL of NaOH added (the half equivalence point), what substances are present in the reaction flask?

14

HA

only

A lot of HA and a little A^-

A lot of A^- and a little HA

A^- only

[HA] = [A^-]

HA + OH^- 🡪

HOH + A^-

15

HA

only

A lot of HA and a little A^-

A lot of A^- and a little HA

A^- only

[HA] = [A^-]

16

HA

only

A lot of HA and a little A^-

A lot of A^- and a little HA

A^- only

[HA] = [A^-]

HA

only

So at the half-equivalence point, pH = pK_a.

​

We can determine the K_a value of a weak acid experimentally using the half-equivalence point of a titration!

A lot of HA and a little A^-

A lot of A^- and a little HA

A^- only

[HA] = [A^-]

HA

only

What is the pH at the half-equivalence point?

pH = 4.8

pH = pK_a = 4.8

pK_a = -log K_a

4.8 = -log K_a

K_a = 1.58x10^-5

A lot of HA and a little A^-

A lot of A^- and a little HA

A^- only

[HA] = [A^-]

Now we can find the K_a value on a table and identify the acid!

What is the pH at the half-equivalence point?

pH = 4.8

pH = pK_a = 4.8

pK_a = -log K_a

4.8 = -log K_a

K_a = 1.58x10^-5

Now we can find the K_a value on a table and identify the acid!

What is the pH at the half-equivalence point?

pH = 4.8

pH = pK_a = 4.8

pK_a = -log K_a

4.8 = -log K_a

K_a = 1.58x10^-5