Answer to Question #145773 in Organic Chemistry for Ali cnd

Question

When 0.08 moles of NaOH into 500 ml of 0.6 M buffer system at pH 6.6, the ratio of [Salt]/[Acid] becomes 0.1 (adding NaOH does not change volume). What is the pKa of the buffer system?

Answer 1

Amount of buffer = 0.6M × 500/1000L = 0.3 mol

When 0.08 moles of NaOH reacts with 0.3 moles of buffer (let's assume the buffer is mono protic)

NaOH is the limiting reagent and the amount of buffer left is 0.3 - 0.08 = 0.22 moles

Concentration of NaOH unreacted = 0.22/500ml = 0.44M

Comparing with the ratio above,

Salt/Acid = 0.1

Since salt = 0.08, acid = 0.8 moles = 0.88/500ml = 1.76M

from,

pKa = pH - log$\frac{[Acid]}{[Conjugate\ base]}$

pKa = 6.6 - log$\frac{1.76}{0.44}$

pKa = 6.6 - 0.602 = 5.998