Additional Aspects of aqueous balances:
2. Determine the pH of a buffer solution consisting of 0.50M of CH3COOH and 0.50M of CH3COO- . Ka=1.8x10^-5
Concentration of CH3COOH = 0.50 M
Concentration of CH3COO- = 050 M
Acid dissociation constant Ka = 1.8 x 10^-5
pKa = - log Ka = - log (1.8 x 10^-5) = 4.74
From the Henderson-Hasselbalch equation
pH = pKa + log [(base)/(acid)]
pH = pKa + log [(CH3COO-)/(CH3COOH)]
pH = 4.74 + log (0.50/0.50)
pH =4.74 + 0
pH = 4.74
Additional Aspects of aqueous balances: 2. Determine the pH of a buffer solution consisting of 0.50M...
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