Q.

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

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a

4.97

b

4.70

c

7.04

d

3.40

answer is A.

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Detailed 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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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