Page 40 - Instrumen Soal HOTS
P. 40
maka pH = 14 – pOH
pH =14–(5−log √5)
pH = 9 + 𝐥𝐨𝐠 √𝟓 = 9 + 2,236 = 11,236
•
Diketahui : n CH3COOH
Percobaan 3
2
= 50 mL x 0,2 M = 10 mmol
n NaOH = 50 mL x 0,2 M = 10 mmol
Ka CH3COOH = 10-5 Vtotal = 100 mL
Ditanya :
pH CH3COONa ...... ?
CH3COOH + NaOH → CH3COONa + H2O
Awal: 10 mmol Akhir: 10 mmol
Sisa: -
10 mmol 10 mmol
-
- - 10 mmol 10 mmol
10 mmol 10 mmol
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