Solution-
The reaction can be written as
CH3COONa + H2O -------> CH3COOH + OH- + Na+
0.0140 - X X X
Now the pKa for CH3COONa = 4.75
So the Kb = 10^-pKb = 1.78*10^-5
=>Kb = 1.78*10^-5 = X^2 / 0.0140 - X
=>1.78*10^-5*0.0140 + 1.78*10^-5*(-x)= x^2
=> X^2+1.78*10^-5x -2.492e-7=0
=> X = 0.000490 M = [OH-]
pOH = - log [OH-] = - log (0.000490)
= 3.30
pH = 14 - 3.30
= 10.7
2- What is the pH for a 0.0140 M solution of sodium acetate (NaAc). The pKa...
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