Calculate the moles in each reactant
Acetic Acid (.231 mol/L)(0.125L)= 2.887 x 10^-2 mol AA
NAOH (.209mol/L)(.037L)= 7.733 x 10^-3 mol NaOH
As you can see, we have more AA and less NaOH, so NaOH is the limiting reactant.
NaOH + CH3COOH -> NaCH3COOH + H2O
start: 4.5 mmol, 8.75mmol, 0, n/a
change: -4.5 mmol, -4.5mmol, 4.5mmol, n/a
after rxn: 0, 4.25 mmol, 4.5mmol, n/a
Now your HH equation
pKa + log(conjugate base/acid) = 4.75 + log(4.5x10^-3 / 4.25x10^-3) =
pH = 4.76
The Henderson-Hasselbalch equation in chemistry and biology connects the pH of a weak acid solution to its acid dissociation constant, Ka, and the ratio of the concentrations of the acid and its conjugate base in equilibrium.
When both the pH of the solution and the ratio of ionized to unionized forms are known, the Henderson Hasselbalch equation can be used to calculate the pKa.
To learn more about Henderson Hasselbalch equation, refer to:
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