Calculate the pH of a 0.800 M NaCH3CO2 solution. Ka for acetic acid, CH3CO2H, is 1.8...

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Calculate the pH of a 0.800 M NaCH3CO2 solution. Ka for aceticacid, CH3CO2H, is 1.8 Ã— 10-5. If someone could list the steps andcalculations that would be helpful. Thank you!

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3.9 Ratings (620 Votes)

NaCH3CO2Â Â it is the salt of strong base and weak acid . so anion participate in hydrolysis process

CH3CO2- + H2O ------------------------> CH3COOH + OH-

0.800Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0 -------------------> initial

-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x ---------------------> changed

0.8-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x -----------------------> equilibrium

Kb = [CH3COOH][OH-]/[CH3COO-]

Kw / Ka = x^2 / 0.8 -x

1.0 x 10^-14 / 1.8 x 10^-5 = x^2 / 0.8 -x

5.56 x 10^-10 = x^2 / 0.8 -x

x^2 + 5.56 x 10^-10 x - 4.45 x 10^-10 = 0

x = 2 .1 x 10^-5

[OH-] = 2 .1 x 10^-5 M

pOH = -log[OH-] = -log (2 .1 x 10^-5)

pOH = 4.68

pH + pOH = 14

pH = 9.32

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