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Pertanyaan

Hitung ph larutan natrium asetat 0,1 m dengan tetapan kesetimbangan 0, 000000009

1 Jawaban

  • CH3COONa= 0,1 M
    Ka = 9 × 10^-9
    pH...???
    [H+] = akar ( 9 × 10^-9 × 10^-1)
    [H+] = akar ( 9 × 10^-10)
    [H+] = 3 × 10^-5
    pH = -log[H+]
    = -log(3×10^-5)
    = 5 - log3

    mohon koreksi ulang, semoga membantu.

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