A 25.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of acetic acid is 4.5E-4
A 25.0 mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. What is the pH at the equivalence point? The Ka of acetic acid is 4.5E-4
note that the actual value of Ka of acetic acid is 1.8*10^-5, not 4.5*10^-4
mmol = MV = 25*0.15 = 3.75 mmol of acetic acid
Vbase = 3.75/0.150 = 25 mL
total V = 25+25 = 50 mL
in equivalance:
CH3COO- + H2O <-> CH3COOH + OH-
Kb = [CH3COOH][OH-]/[CH3COO-]
Kb = Kw/Ka = (10^-14)/(1.8*10^-5) = 5.555*10^-10
5.555*10^-10 = x*x/(M-x)
M = mmol of CH3COO- / total V = 3.75 / 50 = 0.075 M
5.555*10^-10 = x*x/(0.075 -x)
X = [OH-] = 6.45*10^-6
pOH= -log(6.45*10^-6) = 5.190
pH = 14-5.190 = 8.81
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