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


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