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chapter18 - Homework Chapter 18 18.8(a CH3CHOHCO2H(aq H2O(l...

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Homework Chapter 18 18.8 (a) CH 3 CHOHCO 2 H(aq) + H 2 O( l ) ' H 3 O + (aq) + CH 3 CHOHCO 2 (aq) = 2 3 0491 . 0 500 . 0 / 1 . 112 75 . 2 CHOCO CH M L mol g g [ ][ ] [ ] [ ][ ] [ ] [ ][ ] [ ] [ ] [ ] 54 . 3 log 10 9 . 2 100 . 0 0491 . 0 10 4 . 1 100 . 0 0491 . 0 10 4 . 1 3 4 3 4 4 2 3 2 3 3 = = × = = = × + = × = + + + O H pH M O H x x x x x H CHOCO CH CHOCO CH O H K a (b) The buffer solution has a higher pH than the original lactic acid solution (pH = 2.43) because a weak base (CH 3 CHOHCO 2 ) was added to the lactic acid solution. 18.12 p K a = –log( K a ) = –log(5.6 × 10 –10 ) = 9.25 pH = p K a + log [NH 3 ] [NH 4 + ] = 9.25 + log 0.045 0.050 = 9.21 18.18 [H 3 O + ] = 10 –pH = 10 –9.5 = 3.16 × 10 –10 M [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] + + + + = × = × = 4 3 4 3 10 10 4 3 3 8 . 1 10 16 . 3 10 6 . 5 NH NH NH NH NH NH O H K a A solution in which the NH 3 concentration is approximately twice that of NH 4 Cl will have a pH of 9.5. For example, add 2 mol NH 3 and 1 mol NH 4 Cl to some amount of water.
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18.26 (a) M L mol M NH H C mol HCl mol mol L M HCl mol NH H C 180 . 0 0250 . 0 00449 . 0 1 1 00449 . 0 02567 . 0 175 . 0 2 5 6 2 5 6 = = = = = (b) C 6 H 5 NH 3 + (aq) + H 2 O( l ) ' H 3 O + (aq) + C 6 H 5 NH 2 (aq) Total volume = 0.02567 L + 0.0250 L = 0.0507 L 5 10 14 2 5 6 10 5 . 2 10 0 . 4 10 0 . 1 0887 . 0
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