Q. 464.5( 14 Votes )

The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Answer :

Given:

Ionization constant of acetic acid = 1.74 × 10–5


0.05 M solution


Ionization of CH3COOH


CH3COOH CH3COO + H+


By applying the formula,


Ka =


Ka =


At equilibrium [CH3COO ]= [H+]


Hence,


Ka=


As Ka = 1.74 × 10–5 (given)


[CH3COOH] = 0.05 M


1.74 × 10–5 =


[H+]2 = 1.74 × 10–5 × 5 × 10-2M


[H+] =


[H+] = 9.33 × 10-4 M


As, [CH3COO] = [H+]


Thus, the concentration of acetate ion is 9.33 × 10-4M


To calculate pH, we apply the formula,


pH = -log[H+]


pH = -log(9.33 × 10-4)


pH = -log 9.33 – log 10-4


pH = - log 9.33 – (-4) log 10


pH = 4 - log 9.33


pH = 4 – 0.9699


pH = 3.03


Thus, its pH is 3.03.


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