Chapter%2011

Chapter%2011 - 11.4 (a) The reaction is HCN(aq) + H 2 O(l)...

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11.4 (a) The reaction is 2 3 HCN(aq) H O(l) H O (aq) CN (aq). + + + Concentration 1 (mol L ) HCN(aq) + H 2 O(l) H 3 O + (aq) + CN (aq) initial 0.075 0 0.060 change x + x + x equilibrium 0.075 x x 0.060 + x K a = ] [ ] ][ [ 3 HCN CN O H + = ) 075 . 0 ( ) 060 . 0 )( ( x x x + = 4.9 x 10 -10 x = [[H 3 O + ] 6.1 x 10 -10 mol . L -1 (b) 0.30 M NaCl will have no effect: Concentration 1 (mol L ) NH 2 NH 2 (aq) + H 2 O(l) NH 2 NH 3 + (aq) + OH (aq) initial 0.20 0 0 change x + x + x equilibrium 0.20 x x x 2 3 b 2 2 2 6 7 4 1 [NH NH ][OH ] [NH NH ] [OH ] 1.7 10 0.20 [OH ] 3.4 10 5.8 10 mol L + = × = = × = × K 14 11 1 w 3 4 1.00 10 [H O ] 1.7 10 mol L [OH ] 5.8 10 + × = = = × × K (c) Setup is similar to part (a). 10 3 a 10 3 10 1 3 [H O ][CN ] ( )(0.030 ) 4.9 10 [HCN] (0.015 ) [H O ][0.030] 4.9 10 (0.015) [H O ] 2.5 10 mol L + + + + = × = = × = × x x K x
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(d) When the concentrations of a weak base and its conjugate acid are equal, the pOH equals the p K b . Therefore, the pOH of hydrazine = p K b = 5.77, and 8.23 9 1 3 pH 14.00 pOH 14.00 5.77 8.23 [H O ] 10 5.9 10 mol L + = = = = = × 11.12 HA = acetylsalicylic acid A = conjugate base of acetylsalicylic acid 2 3 HA(aq) H O(l) H O (aq) A (aq) + + + 4 a 3.2 10 = × K We will use the following formula to evaluate the ratio of [A ] to [HA] at equilibrium: a [A ] pH p log [HA] = + K The formula is derived in this way: 3 a a 3 [H O ][A ] [HA] [HA] [H O ] [A ] + + = = K K a a a [HA] pH p [A ] [HA] p log [A ] [A ] p log [HA] = = = + K K K Given that pH = 4.13 and K a = 3.2 x 10 -4 (pKa = 3.49) 4.13 = 3.49 + log ] [ ] [ HA A ] [ ] [ HA A = 4.37
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11.36 (a) It is a weak base because the stoichiometric point pH
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Chapter%2011 - 11.4 (a) The reaction is HCN(aq) + H 2 O(l)...

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