Biologi

Pertanyaan

100 mL larutan ca(OH)2 0,1 m dicampurkan ke dalam larutan CH3COOH 0,1 M,ternyata PH campurannya =5. jika harga Ka asam asetat 1x10^5,maka volume larutan CH3COOH 0,1 adalah..

1 Jawaban

  • dik : pH = 5
    [H+] = 10^-5
    Ca(OH)2= 100ml × 0,1 M = 10 mmol
    Ka = 10^-5
    CH3COOH = 0,1 M
    dit : Volume CH3COOH
    jawab :
    [H+] = Ka×Al/Bk
    10^-5 = 10^-5×Al/10
    10^-4= Al×10^-5
    Al = 10^(-4-(-5)
    Al = 10 mmol

    n Al = v × M
    10 mmol = v × 0,1
    v = 100 ml

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