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3. (11) if you add 0.075 moles of acetic acid (K1.8 X 10), and 4.15 g of NaA (MM-83 /mol) to a single flask, and then add an
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moles of acetic acid = 0.075 mol 9 of acetic acid = 1.8xio 5 mass of daaC = 4.159 (MMormolar mass of NAC = 83 g/mol total vol= 0.05 mol 0.125 mol = 0 1225M. 0.41 molarity of Na Aa = 1o. 125 M . dow, we know that . Pka = -log(ka) = -log(1.8 x 10 Pka =

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