Kimia

Pertanyaan

Hitunglah ph larutan CH3COOH 0.05 M dgn ka= 1,8×10-5

1 Jawaban

  • CH3COOH 0.05 (Ka=1,8 x 10-5)
    asam lemah
    [H+]= √Ka x Ma
    = √18 x 10-6 x 5 x 10-2
    = √90 x 10-8
    = √9 x 10-7
    = 3√10 . 10-4
    ph = -log [H+]
    = -log [3√10 . 10-4
    ph = 4 - log 3√10

Pertanyaan Lainnya