Pertanyaan

Berapa ml volume Ba(OH) 2 0,1 M dan larutan CH 3 COOH 0,05 M berturut-turut agar diperoleh 400 ml larutan penyangga dengan pH = 5 + 2 log 2? (Ka CH 3 COOH = 10 -5 )

Berapa ml volume Ba(OH)₂ 0,1 M dan larutan CH₃COOH 0,05 M berturut-turut agar diperoleh 400 ml larutan penyangga dengan pH = 5 + 2 log 2?

(Ka CH₃COOH = 10^-5)

18 mL dan 382 mL
19 mL dan 381 mL
20 mL dan 380 mL
21 mL dan 379 mL
22 mL dan 378 mL

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Pembahasan

pH = 5 + 2 log 2 = 5 + log 2 2 = 5 + log 4, sehingga [H + ] = 4 x 10 -5 Total larutan buffer = 400 mL, Maka jika volume Ba(OH) 2 = v, maka volume CH 3 COOH = (400 – v) Mol Ba(OH) 2 = M x V = 0,1 M x v mL = 0,1v mmol Mol CH 3 COOH = M x V = 0,05 M x (400 – v) mL = 20 – 0,05v mmol Maka di dapat v = 19 mL Volume Ba(OH) 2 = v = 19 mL Volume CH 3 COOH = (400 – v) = 400 – 19 = 381 mL

pH = 5 + 2 log 2 = 5 + log 2² = 5 + log 4, sehingga [H⁺] = 4 x 10^-5

Total larutan buffer = 400 mL,

Maka jika volume Ba(OH)₂ = v, maka volume CH₃COOH = (400 – v)

Mol Ba(OH)₂ = M x V = 0,1 M x v mL = 0,1v mmol

Mol CH₃COOH = M x V = 0,05 M x (400 – v) mL = 20 – 0,05v mmol

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Maka di dapat v = 19 mL

Volume Ba(OH)₂ = v = 19 mL

Volume CH₃COOH = (400 – v) = 400 – 19 = 381 mL

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