° LARUTAN PENYANGGA °

Diketahui

• M = 0,1 M
• V = 100 mL
• pH = 3
• m NaOH = 0,4 g

Ditanya

• pH = ?

Jawab

1. Tentukan mol awal asam asetat

n = M • V

n = 0,1 • 100

n = 10 mmol

2. Tentukan Ka larutan asam asetat

pH = - log [H⁺]

3 = - log [H⁺]

[H⁺] = 10⁻³

[H⁺] = √(Ka • M)

Ka • 0,1 = 10⁻⁶

Ka = 10⁻⁵

3. Tentukan diagram reaksi

R : CH3COOH + NaOH → CH3COONa + H2O

a : 10 10

r : 10 10 10 10

s : 0 0 10 10

4. Tentukan nilai konsentrasi basa konjugasi

M CH3COONa = n / V

M CH3COONa = 0,01 mol / 0,1

M CH3COONa = 0,1 M

5. Tentukan nilai pH larutan

[OH⁻] = √(Kw / Ka • M)

[OH⁻] = √(10⁻¹⁴ / 10⁻⁵ • 0,1)

[OH⁻] = 10⁻⁵

pOH = - log [OH⁻]

pOH = - log[10⁻⁵]

pOH = 5

pH = 14 - pOH

pH = 14 - 5

pH = 9