Solutions_Manual_for_Organic_Chemistry_6th_Ed 18

Solutions_Manual_for_Organic_Chemistry_6th_Ed 18 - 1 -13 1...

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Unformatted text preview: 1 -13 1 mole HEr (a) 5 .00 g HBr x 80.9 g HBr 0.06 1 8 moles HEr 0.0618 moles = 0 .06 1 8 moles HBr 0.06 1 8 moles H 30 + (100% dissociated) x = 1 00 mL - H30 + 1000 1L mL = 0.6 1 8 moles H30 + 1 L solution = pH = 10gIO [H30+] - logl o (0.6 1 8) � (b) 0 .0375 moles NaOH 1 .50 g NaOH x 1 mole NaOH 40.0 g NaOH 0.0375 moles -OH (100% dissociated) 0.0375 moles NaOH = 0.0375 moles -OH 50. mL = 1 000 mL 0.75 moles -OH 1 L solution lL 14 1 x 1 0-14 1 .33 x 10-14 [H 3 0+] 1 x 1 0- = 0.75 [-OH] pH -loglO [H30+] -logl o (1.33 x 10- 14 ) x = = = = = 0 .75 M (the number of decimal places in a pH value is the number of significant figures) 1-14 (a) By definition, an acid is any species that can donate a proton. Ammonia has a proton bonded to nitrogen, so ammonia can be an acid (although a very weak one). A base is a proton acceptor, that is, it must have a pair of electrons to share with a proton; in theory, any atom with an unshared electron pair can be a base. The nitrogen in ammonia has an unshared electron pair so ammonia is basic. In water, ammonia is too weak an acid to give up its proton; instead, it acts as a base and pulls a proton from water to a small extent. (b) water as an acid: H 2 O + NH3 -OH + NH4+ -- H 2 O + H Cl water as a base: (c) methanol as an acid: CH30H + NH3 methanol as a base: CH30H + H 3 O+ C lCH3O+ H2 SO4 4+ CH3OH2 + + H S04+ NH 8 ...
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