Calculate the pH of a solution of 0.60 M of NaC2H3O2 (Ka for HC2H3O2 = 1.8 x 10^

Question

Calculate the pH of a solution of 0.60 M of NaC2H3O2 (Ka for HC2H3O2 = 1.8 x 10^-5)
PH = _________________

Explanation / Answer

Ka of CH3COOH = 1.8 x 10^-5 CH3COONa is a strong salt : >> CH3COO- + Na+ acetate ion reacts with water : CH3COO- + H2O CH3COOH + OH- the concstant of this equilibrium also called hydrolisis constant is Kh = Kw / Ka = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.6 x 10^-10 5.6 x 10^-10 = x^2 / 0.60-x x = [OH-] = 1.83 x 10^-5 M pOH = - log [OH-] = 4.7375 pH = 14 - 4.7375 = 9.2625 NOTE : sodium actetate is a salt derived by a strong base ( NaOH ) and a weak acid ( acetic acid). . This means that it is a basic salt and pH must be > 7

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