Given:

Degree of ionization (α)= 0.132

Concentration(C) = 0.1 M

Ionization of bromoacetic acid

By applying the formula

K_a =

⇒K_a =

⇒K_a=

We can neglect C(1-α) as its value is very small

∴ K_a = Cα²

As α= 0.132, C = 0.1 M (Given)

Hence,

⇒K_a = 0.1× 0.132 × 0.132

⇒K_a = 1.74 × 10^-3

To calculate Pk_a , we apply the formula:

Pk_a = -logK_a

As K_a = 1.74 × 10^-3

∴ pk_a = -log ( 1.74 × 10^-3)

⇒Pk_a = -log 1.74 – (-3) log 10

⇒Pk_a = -0.2405 + 3

⇒Pk_a = 2.76

Thus, the pK_a of bromoacetic acid is 2.76

[H⁺] = Cα

⇒[H⁺] = 0.1 × 0.132

⇒[H⁺] = 1.32 × 10^-2

To calculate pH, we apply the formula:

pH= -log[H⁺]

⇒pH= -log (1.32 × 10^-2)

⇒pH= -log 1.32 + 2log 10

⇒pH = 1.88

Thus, the pH of the solution is 1.88