Page 78 - KIMIA SMA KELAS XI BERBASIS PjBl

3.
Diketahui : V = 100mL Ka = 1 × 10−5
M = 0,1 M Ditanya : pH ....?
Jawab :
CH3COOH 0,1 M ( Ka = 10-5 )
Jawab : [𝐻+] = √𝐾𝑎 × 𝑀 [𝐻+] = √10−5 × 0,1
[𝐻+] = √10−6 [𝐻+] = 10−3
4.
𝑝𝐻 = −𝐿𝑜𝑔[𝐻+] 𝑝𝐻 = −𝐿𝑜𝑔[10−3]
𝑝𝐻 = 3
Diketahui : V = 100mL M = 0,2 M
pH = 11 + Log 2 Ditanya : Kb ...?
Jawab
Mencari nilai [OH-]
pH = 11+Log 2 , maka
pH + pOH = 14 (11+Log2) + pOH = 14 pOH = 14 – ( 11 + Log 2 ) pOH = 3 - Log 2
pOH = - Log [OH-]
[OH-] = 2 x 10-3
Indicator
MO Kuning
BTB Kuning
MM Jingga
PP Tidak berwarna
Mencari Nilai Kb
[OH−] = √Kb × M
2 x 10−3 = √Kb × 0,2
2 x 10−3 = √Kb × 20 × 10−2 2 x 10−3 = 4,47 × 10−1√Kb
√Kb= 2x10−3 4,47 × 10−1
√Kb = 0,447 × 10−2
𝐾𝑏 = √0,447 × 10−2 𝐾𝑏 = 0,668 × 10−1 𝐾𝑏 = 6,68 × 10−2
Hasil Analisis
pH > 4,4
pH < 6,0
4,4 ≤ pH ≤ 6,2 pH < 8,3
5.
Data Hasil
Berdasarkan hasil analisis tersebut, maka pH air sungai tersebut adalah 4,4 ≤ pH ≤6,0
Kimia Kelas XI SMA/MA
68