Question

The degree of dissociation of 0.1 M CH3COOH in 0.1 M HCl will be {Ka(CH3COOH) = 10 ^-5}

1) 10^-3 2)10^-4

Answer 1

in the presence of 0.1M HCl, Let us ‘y’ is the amount of acetic acid dissociates.

at equilibrium ,

[CH3COOH] = 0.01 - y ≈ 0.01M [ as y << 0.01 , because it is weak acid ]

[CH3COO-] = y

[H^+] = 0.1 + y [ as concentration of HCl is 0.1M so, already 0.1M of H^+ is present and at equilibrium it increases with y ]

now, Ka = [CH3COO^-][H^+]/[CH3COOH]

or, 10^-5 = y(0.1)/(0.01)

or, y = 10^-6

so, degree of dissociation is 10^-6

and percentage of degree of dissociation is 10^-4 %