Question

Calculate the percent ionization of 1.45 M aqueous acetic acid solution. K₂ for acetic acid is 1.8x10.

(A) 0.35%

(B) 0.0018%

(C) 0.29%

(D) 0.0051%
​

Answer 1

Answer:

ka = [H+]² / 0.1

Ka = 1.8*10^-5

1.8*10^-5 = [H+]² / 0.1

[H+]² = (1.8*10^-5)* 0.1

[H+]² = 1.8*10^-6

[H+] = 0.0013

Explanation:

Mark me as brainliest

Answer 2

Answer:

Option A is the correct answer

Explanation:

Please mark ne as the BRAINLIEST