Kimia

Pertanyaan

Larutan ch3coona sebanyak 10 ml memiliki ph=8. Jika ka ch3cooh=1.10-5 garam ch3coona yang terlarut dalam 200 mL larutannya sebanyak (Ar. C=12 g mol H=1 g mol O=16 g mol Na=23 g mol

1 Jawaban

  • pH = 8

    pOH= 14 - pH = 14 - 8 = 5

    [OH-] = 10^-5

    CH3COONa -> CH3COO- + Na+

    [OH-] = √kw/ka x [CH3COO-]

    10^-5 = √10^-14/10^-5 x [CH3COO-]

    10^-10 = 10^-9 x [CH3COO-]

    [CH3COO-] = 10^-10/10^-9 = 10^-1 M

    [CH3COONa] = 1/1 x 10^-1 = 10^-1 M

    M = massa/Mr x 1000/V

    10^-1 = massa/70 x 1000/200

    massa = 1,4 gram

Pertanyaan Lainnya