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Chemistry

Calculate the ph of 0.125M of NaAc

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4.4 Ratings (661 Votes)

The give compound is a salt of sodium acetate (CH3COO-Na+)

The Ka of acetic acid = 1.8 X 10^-5

So Kb of acetic acid = Kw / Ka = 10^-14 / 1.8X 10^-5 = 5.56 X 10^-10

It will dissociate as

CH3COONa => CH3COO- + Na+

Now

             CH3COO- + H2O <------> CH3COO- + OH-

Initial          0.125                             0             0

Change        -x                                   x              x

Equilibrium 0.125-x                       x                  x

Kb = x^2 / (0.125-x)

x <<1 so will ignore in denominator

1.56 X 10^-10 = x^2 / 0.125

x^2 = 0.195 X 10^-10

x = 0.442 X 10^-5 = 4.42 X 10^-6 = [OH-]

pOH = -log[OH-] = -log[4.42 X 10^-6]

pOH = 5.35

pH = 14-5.35 = 8.65


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