Henderson-Hasselbalch equation?
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Using the Henderson-Hasselbalch equation (pH = pKₐ + log₁₀ [salt/acid]) for a buffered solution, what is the apparent pKₐ of the acid at room temperature resulting from the addition of 5 mL of 0.2 M sodium hydroxide to 10 mL of 0.2 M acetic acid and having a final hydrogen ion concentration of 1.86 x 10-5 M? (log₁₀ 1.86 = 0.2695)
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0.20 moles sodium hydroxide/liter x 0.005 liters = 0.001 moles acetic acid
NaOH + HC2H3O2 ==⇒ NaC2H3O2 + H2O
0.20 moles acetic acid x 0.010 liters = 0.002 moles acetic acid
0.0010 moles of NaOH will react with 0.0010 moles of acetic acid producing 0.0010 moles of C2H3O2- and leaving 0.001 moles of acetic acid.
[HC2H3O2] = 0.0010 moles /0.015 liters = 0.0666
[C2H3O2-] = 0.0010 moles/0.015 liters = 0.0666
[H+] = 1.8 x 10^-5 ;
pH = -log[H+] = -log(1.8 x 10^-5) = -(-4.73) = 4.73
pH = pKa + log[C2H3O2-]/[HC2H3O2]
4.73 = pKa + log(0.0666)/0.0666
4.73 = pKa + log(1) since log 1 = 0
pKa = 4.73