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 0.09 M ${\mathrm{CH}}_{3}\mathrm{COOH}$ and 0.15 M of potassium acetate?

1. A

4.97

2. B

4.70

3. C

7.04

4. D

3.40

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### Solution:

${\mathrm{P}}^{\mathrm{H}}={\mathrm{P}}^{{\mathrm{K}}_{\mathrm{a}}}+\mathrm{log}\frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}$

$=+4.7447+0.699-0.4771$

$=4.9666$

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