Jawaban:

C. 10,5

Penjelasan:

Diketahui:

• V awal = 100 mL = 0,1 L = 10⁻¹ L

• pH awal = 11
• Kb = 10⁻⁵
• V₂ = 100 mL + 900 mL = 1.000 mL = 1 L

Ditanya:

• pH akhir

Jawab:

A. Cari [OH⁻] awal

pH = 11

→ pOH = 14 - 11

pOH = 3

pOH = -log[OH⁻]

3 = -log[OH⁻]

[OH⁻] = 10⁻³

B. Cari konsentrasi (M) awal

[OH⁻] = √(Kb × M awal)

10⁻³ = √(10⁻⁵ × M awal) → Kuadratkan kedua ruas

10⁻⁶ = 10⁻⁵ × M awal

M awal = 10⁻⁶ / 10⁻⁵

M awal = 10⁻¹

C. Cari jumlah mol (n)

M awal = n / V awal

→ n = M awal × V

n = 10⁻¹ × 10⁻¹

n = 10⁻²

D. Cari konsentrasi (M) akhir/setelah pengenceran

M akhir = n / V akhir

M akhir = 10⁻² / 1

M akhir = 10⁻²

E. Cari [OH⁻] akhir/setelah pengenceran

[OH⁻] = √(Kb × M akhir)

[OH⁻] = √(10⁻⁵ × 10⁻²)

[OH⁻] = √10⁻⁷

[OH⁻] = 10⁻³,⁵

F. Cari pH akhir/setelah pengenceran

pOH = -log[OH⁻]

pOH = -log(10⁻³,⁵)

pOH = 3,5

pH = 14 - pOH

pH = 14 - 3,5

pH = 10,5

Jadi, jawaban yang benar adalah C. 10,5