1aa3-2012-tut1-a

1aa3-2012-tut1-a - CHEMISTRY 1AA3 Week of JANUARY 9 2012...

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CHEMISTRY 1AA3 Week of JANUARY 9, 2012 TUTORIAL PROBLEM SET 1 – Full SOLUTIONS ______________________________________________________________________________ Chem 1AA3 – Tutorial 1 - Full Solutions Page 1 of 10 Assume all calculations are at 25ºC, so that K w = 1.0 × 10 -14 (and pK w = 14.00). 1. What is the conjugate acid of each of the following species? Base Conjugate acid (a) ClO 2 H C l O 2 (b) SO 3 2 H S O 3 2. What is the conjugate base of each of the following species? A c i d C o n j u g a t e b a s e (a) NH 4 + N H 3 (b) NH 3 N H 2 (c) HS - S 2 3. Write balanced chemical reactions and K a or K b expressions for the ionization of the following acids and bases. (a) HNO 2 (b) CH 3 COOH (c) (CH 3 ) 3 N (a) HNO 2 + H 2 O NO 2 + H 3 O + K a = [NO 2 ][ H 3 O + ] [HNO 2 ] (b) CH 3 COOH + H 2 O H 3 O + + CH 3 COO K a = [H 3 O + ] [CH 3 COO ] [ C H 3 COOH] (c) CH 3 NH 2 + H 2 O CH 3 NH 3 + + OH K b = [CH 3 NH 3 + ] [OH ] [ C H 3 NH 2 ] 4. (a) What is the pH of a 10.0 M solution of HCl? (b) What is the pH of a 1.00 × 10 8 M solution of HCl? SOLUTION: (a) pH = log 10 [H 3 O + ] = log 10 (10.0) = 1.000 Very acidic! The point of this question is just to show that the pH scale doesn't just go from 0 to 14, it's just that most of the substances we talk about lie within that range. (3 sig figs Æ 3 decimal places)
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Chem 1AA3 – Tutorial 1 – Full Solutions Page 2 of 10 (b) Again, pH = log 10 [H 3 O + ] BUT in this case, what is [H 3 O + ]? Remember that from the autoionization of water, there is a concentration of [ H 3 O + ] = 1.0 × 10 7 M (which we usually ignore in our calculations). In this case, however, the concentration is very significant. Thus, we have two sources of H 3 O + in our solution: the H 2 O and the HCl: [H 3 O + ] = 1.00 × 10 8 M (from HCl) + 1.0 × 10 7 M (from water) = 1.1 × 10 7 M But, in order to maintain the K w equilibrium there will be a small shift to recombine so [H 3 O + ] = 1.1 × 10 7 – x and [OH - ] = 1.0 × 10 7 – x, solving for x = 0.05 × 10 7 M. Normally this shift is very small relative to the initial amount of the larger concentration. Now we can use pH = log 10 [H 3 O + ] = log 10 (1.0 5 × 10 7 ) = 6.98 If you ignore the [H 3 O + ] from water in this case, the pH will work out to be 8, which doesn't make sense, since we're talking about a dilute solution of acid ! If you ignore the shift back to equilibrium, you get pH = log 10 (1.1 × 10 7 ) = 6.96, a small error. 5. Dimethylamine, (CH 3 ) 2 NH, is a weak base (ionization constant K b = 7.40 × 10 4 ). (a) What is the equilibrium concentration of dimethylammonium ion, [(CH 3 ) 2 NH 2 ] + in a 0.400 M aqueous solution of (CH 3 ) 2 NH? (b)
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1aa3-2012-tut1-a - CHEMISTRY 1AA3 Week of JANUARY 9 2012...

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