Jika 50 mL CH3COOH 0,1 M direaksikan dengan 50 mL NaOH 0,1 M menghasilkan garam. Harga pH larutan adalah … (Ka CH3COOH = 10^-5) (soal lebih jelas ada di gambar)
Jawaban:
n CH₃COOH = 50 x 0,1 = 5 mmol = 5 . 10⁻³ mol
n NaOH = 50 x 0,1 = 5 mmol = 5 . 10⁻³ mol
CH₃COOH + NaOH ⇒ CH₃COONa + H₂O
Mula 5 . 10⁻³ 5 . 10⁻³ - -
Reaksi 5 . 10⁻³ 5 . 10⁻³ 5 . 10⁻³ 5 . 10⁻³
Akhir - - 5 . 10⁻³ 5 . 10⁻³
[tex][OH-] = \sqrt{\frac{K_{w} }{K_{a} } M_{a} } = \sqrt{\frac{10^{-14} }{10^{-5} }(5. 10^{-3} ) }\\ \\ = \sqrt{5.10^{-12} } = 10^{-6}\sqrt{5}[/tex]
pOH = - log [OH⁻] = - log 10⁻⁶ √5 = 6 - log √ 5
pH = 14 - pOH = 14 - ( 6 - log √ 5 ) = 8 + log √5
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