Question

PLEASE BANTU!!!!
Massa ch3coona (mr = 82) yang harus ditambahkan ke dalam 1000 ml larutan ch3cooh 0,1 m agar ph-nya 6 adalah.. gram (ka ch3cooh = 10^-5

Answer 1

H+=10^-6
n CH3COOH = 0,1 x 1000 = 100 mmol = 0,1 mol

H+= Ka x (na/ng)
10^-6= 10^-5 (0,1/ng)
ng = 1 mol
n CH3COONa = 1 mol
massa CH3COONa = n x Mr = 1 x 82 = 82 gram

Answer 2

Diket dan dit:
*Mr = 82
*pH = 6 --> [H+] = 10^-6
*n CH3COOH = 100 mmol
*Ka = 10^-5
*Massa garam?

Yang diket garam (CH3COONa) dan asam lemah (CH3COOH), maka memakai rumus buffer:
[H+] = Ka x [CH3COOH] / [CH3COONa]
10^-6 = 10^-5 x 100 mmol / n CH3COONa
n CH3COONa = 1000 mmol atau 1 mol

Maka, massanya:
n = g/Mr
g = 1 mol x 82 gram/mol
   = 82 gram