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
pKa = 6.6 - log
pKa = 6.6 - 0.602 = 5.998
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