Calculate the pH in 0.10 M solutions of NaCH3CO2. (Ka CH3CO2H = 1.8 x 10-5)?

Sodium acetate is the salt of a weak acid. The acetate anion will undergo hydrolysis as follows:

Ac^- + H2O <===> HAc + OH^-

Ac^- is short for CH3CO2^-

We can write a Kb expression for the above equilibrium:

Kb = ([HAc] [OH^-]) / [Ac^-]

To get the Kb for the above reaction, we do this:

Ka * Kb = Kw

(1.8 x 10^-5) (Kb) = 1.0 x 10^-14

Kb = 5.6 x 10^-10

Now, use the Kb expression:

5.6 x 10^-10 = [(x) (x)] / 0.10

x = 7.45386 x 10^-6 M (this is the hydroxide conc, I'll keep a few extra digits for the moment)

The usual path is to determine the pOH and fom that, get the pH. I'll do that:

pOH = -log 7.45386 x 10^-6 = 5.1276

pH = 14 - 5.1276 = 8.8724

To two sig figs, the answer is 8.87 (and, yes, that's 2 sig figs. Only the mantissa contains sig figs in a logarithm.)

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