Question

37. Ke dalam 100 mL larutan CH3COOH 0,2 M (Ka=1x10-5) ditambahkan 100 mL larutan NaOH 0,2 M sehingga terjadi reaksi:

CH3COOH (aq) + NaOH (aq) → CH3COONa (aq) + NaOH (l)

Tentukan harga pH larutan yang terbentuk! (Kw=10-14)

Answer 1

Verified answer

Mol asetat = 100 x 0.2 = 20 mmol
mol NaOH = 100 x 0.2 = 20 mmol

..CH3COOH + NaOH => CH3COONa + H2O
m.......20...............20
r.........-20..............-20................+20.............+20
s.........-....................-........................20..............20

G = mol garam/v total = 20/200 = 0.1

OH- = √kw/ka x G = √10^-14/10^-5 x 0.1
OH- = 10^-5
pOH = 5
pH = 14-5 = 9