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Unformatted text preview: H 2 CO 3 + NaCl b. H 2 CO 3 + NaOH ------> NaHCO 3 + H 2 O 5a. 4.75 = 4.75 + log (0.1/0.1) b. 5.75 = 5.75 + log (0.1/0.01) c. 6.37 = 6.37 + log (0.1/0.1) d. 7.37 = 7.37 + log (0.1/0.01) The calculate values were relatively close. Where there were discrepancies most likely is error in the lab preparation....
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