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How would you prepare 1.20 L of a buffer with a pH of 9.35 from 0.200 M Na2CO3 and 0.450 M HCl? (This problem requires valuesPlease give the answers in mL. The Ka value given by my textbook for carbonic acid are: Ka1=4.45x10^-7 and Ka2=4.69x10^-11.

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Answer #1

total = 102 Lit = 12oome (Na, (03)= 0.2M [HC] = 0.45m VNU 2CO3 = vizit VHCI = (1.2 - vi) Lit nNa.CO3 = 60.2 xvi) mol nHCl - (+ Nazcoz HCI - NUHCO3 + H2O mol initial 0.220, 00-54 -ousui) - - final (0.65 U, -o-su) - (0-54-0-usun) - CNQZ (03] ply = pka,

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