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Dalam 100 mL larutan , terlarut 3,6 gram natrium benzoat (Mr = 144). Ka asam benzoat sebesar 6 x10-5 . Hitung pH larutan .
Dalam 100 mL larutan , terlarut 3,6 gram natrium benzoat (Mr = 144). Ka asam benzoat sebesar 6 x10-5 . Hitung pH larutan .
Konsentrasi natrium benzoat = 0,025/ 0,1 = 0,25 M
[H⁺] = √Ka . M
[H⁺] = √6.10⁻⁵ . 0,25
[H⁺] = √15.10⁻⁶
[H⁺] = 3,87.10⁻³
pH = -log [H⁺]
pH = -log 3,87.10⁻³
pH = 3 - log 3,87