The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.
Given:
Degree of ionization (α)= 0.132
Concentration(C) = 0.1 M
Ionization of bromoacetic acid
By applying the formula
Ka =
⇒Ka =
⇒Ka=
We can neglect C(1-α) as its value is very small
∴ Ka = Cα2
As α= 0.132, C = 0.1 M (Given)
Hence,
⇒Ka = 0.1× 0.132 × 0.132
⇒Ka = 1.74 × 10-3
To calculate Pka , we apply the formula:
Pka = -logKa
As Ka = 1.74 × 10-3
∴ pka = -log ( 1.74 × 10-3)
⇒Pka = -log 1.74 – (-3) log 10
⇒Pka = -0.2405 + 3
⇒Pka = 2.76
Thus, the pKa 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