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3 gram of acetic crystal (mr 60) is dissolved in water until the volume is 500 mL. If Ka of the acetic acid is 10^-5, the degree of ionization of acetic acid is.…

1 Jawaban

  • M = 3/60 × 1000/500 = 0.1 M
    Degree of ionization = α
    α = √(Ka/M) = √(10^-5/0.1) = 10^-3 = 0.001

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