Jika diketahui Ka CH3COOH = 1 × 10 - 4 , maka pH campuran dari 100 mL CH3COOH 0,05 M dan 100 mL KOH 0,05 M adalah...

<p><strong>Jawabannya adalah 8 + log 5.</strong></p><p>&nbsp;</p><p>Campuran asam lemah dan basa kuat apabila tepat habis bereaksi maka akan membentuk garam yang mengalami hidrolisis sebagian dan bersifat basa. Contohnya adalah campuran 100 mL CH₃COOH 0,05 M dengan 100 mL KOH 0,05 M.</p><p>Berikut perhitungan pH campurannya.</p><p>&nbsp;</p><p>nCH₃COOH = M. V = 0,05 M . 100 mL = 5 mmol</p><p>nKOH = M. V = 0,05 M . 100 mL = 5 mmol</p><p>&nbsp;</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;CH₃COOH &nbsp; + &nbsp; &nbsp; KOH &nbsp; &nbsp; &nbsp; &nbsp; → &nbsp; &nbsp; &nbsp;CH₃COOK &nbsp; &nbsp; + &nbsp;H₂O</p><p>mula-mula &nbsp; &nbsp; &nbsp; &nbsp;5 mmol &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;5 mmol</p><p>reaksi &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; -5 mmol &nbsp; &nbsp; &nbsp; &nbsp;-5 mmol &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; +5 mmol</p><p>sisa &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;0 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 0 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 5 mmol</p><p>Terbentuk 5 mmol garam CH₃COOK.</p><p>[CH₃COOK] = n/Volume total</p><p>[CH₃COOK] = 5 mmol ÷ (100 + 100) mL</p><p>[CH₃COOK] = 25 . 10^-3M</p><p>&nbsp;</p><p>CH₃COOK → CH₃COO⁻ + K⁺</p><p>CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻&nbsp;</p><p>Dihasilkan ion OH⁻ yang menunjukkan larutan bersifat basa.</p><p>[OH⁻] = √Kw . [CH₃COOK] . valensi/Ka</p><p>[OH⁻] = √10^-14 . 25 . 10^-3. 1/1. 10^-5</p><p>[OH⁻] = √25 . 10^-12</p><p>[OH⁻] = 5 . 10^-6M</p><p>pOH = -log [OH⁻]&nbsp;</p><p>pOH = -log &nbsp;(5 . 10^-6)</p><p>pOH = 6 - log 5</p><p>pH = 14 - pOH</p><p>pH = 14 - (6 - log 5)</p><p>pH = 8 + log 5</p><p>&nbsp;</p><p><strong>Jadi, pH campuran sebesar 8 + log 5.</strong></p>