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Pertanyaan

3 gram of acid 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 lonization of acetic acid is

1 Jawaban

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

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