What is the PHof a buffer containing 0.09 M CH3COOH and 0.15 M of potassium acetate? Ka=1.8×10-5, log 5=0.699, log 3 =0.4771, log 1.8=0.2553

What is the PHof a buffer containing 0.09 M CH3COOH and 0.15 M of potassium acetate? Ka=1.8×10-5, log 5=0.699, log 3 =0.4771, log 1.8=0.2553

  1. A

    4.97

  2. B

    4.70

  3. C

    7.04

  4. D

    3.40

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    Solution:

    PH=PKa+logsaltacid

         =-log1.8×10-5+log0.150.09

         =-0.2553-5+log 5-log 3

         =+4.7447+0.699-0.4771

         =4.9666

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