What is the oh- concentration of a 0.08 m solution of ch3coona?
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CH3COO- + H2O ---------> CH3COOH + OH-
As,
Kw = [ H+ ] [ OH- ]
Ka = [ H + ] [ A - ] / [ HA ]
Kb = [ CH3COOH ] [ OH - ] / [ CH3COO - ]
Kb = Kw / Ka
Kw = 1 e -14
Ka = 1.8 x 10 ^ -5
Kb = 5.6 e -10 = X^2 / 0.08
X = [ OH - ] = 0.67 e-5 M
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