Moles of acetic acid = molarity * volume in liters = 0.11 * 1 = 0.11 moles
Let n is the moles of CH3COONa.
pKa = - log (Ka) = - log (1.8*10^-5) = 4.745
pH = pKa + log [CH3COONa]/[CH3COOH]
4.40 = 4.745 + log (n/0.11)
4.40 - 4.745 = log(n/0.11)
-0.345 = log (n/0.11)
n/0.11 = 10^(-0.345)
n = 0.0497 moles
Mass of CH3COONa = moles * molar mass = 0.0497 * 82.034 = 4.08 grams ... Answer
Let me know if any doubts.
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