Given:

pH of = 4.15

Concentration of HA = 0.01M

To calculate the concentration of the anion, we apply the formula:

pH = -log[H⁺]

To calculate the ionization constant of the acid, we apply the formula:

K_a =

To calculate the pK_a of an organic acid, we apply the formula:

pK_a = -log k_a

where K_a is the ionization constant of the acid

Let the HA be the acid

The ionization of organic acid HA is HA ⇌ H⁺ + A^-

pH = -log[H⁺]

We can also write,

log[H⁺] = -pH

As pH = 4.15 (given)

∴ log[H⁺] = -4.15

By taking antilog of both the sides, we get

[H⁺] = 7.08 × 10^-5

[A^-] = [H⁺] = 7.08 × 10^-5 M

Thus, the concentration of anion is 7.08 × 10^-5M

K_a =

K_a =

As [A^-] = [H⁺] = 7.08 × 10^-5 (calculated above)

[HA] = 10^-2 (given)

∴ K_a =

⇒K_a = 5.0 × 10^-7

Thus, the ionization constant of the acid is 5.0 × 10^-7

pK_a = -log k_a

As K_a = 5.0 × 10^-7

∴ pK_a = -log (5.0 × 10^-7)

⇒pK_a = - log 5 – (-7) log 10

⇒pK_a = 7 - log 5

⇒pK_a = 7 – 0.699

⇒pK_a = 6.301

Thus, pk_a is 6.301

Note: pK_a is a measure of acid strength. It depends on the identity and chemical properties of the acid. pH is a measure of [H⁺] in a solution. For acids, the smaller the pKa, the more acidic the substance is (the more easily a proton is lost, thus the lower the pH).