Question

calculate the p of 0.1 M of Acetic acid solution dissociation constant of Acetic acid is 1.8 × 10^-5 M​

Answer 1

Given:

Normality of acetic acid = 0.1 N (decinormal solution)

Dissociation constant = Ka = 1.8x10-5

Solution:

CH3COOH + H2O ? CH3COO- +H3O+

Ka = [H3O+][CH3COO-]/[CH3COOH]

1.8x10-5 = x2 / (0.1- x)

x is very negligible compare to initial concentration of acid.

Hence,

1.8x10-5 = x2 / 0.1

x2 = 1.8 x 10-5 x 0.1

x =[H3O+] = 0.001342 N

0.001342 N acetic acid = 0.001342 M acetic acid

As [H+] = 0.001342 M

pH = -log10[H+]

pH= - log10(0.001342)

pH= 2.87

Answer 2

Answer in attachment!

Explanation:

• ᴍʀᴋ. ᴛʜᴀɴᴋs ᴀɴᴅ ʙʀᴀɪɴʟɪsᴛ