Exam 2 Key - Chem 10181, Fall 2007 (1) EXAM 2 KEY October...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Chem 10181, Fall 2007 EXAM 2 KEY October 16, 2007 1 (1) (8 pts. each) Below are listed four pairs of molecules. For each pair, indicate which is more acidic, and explain why. H C C H H C N HCN is more acidic than HCCH because the electronegative N helps to stabilize the negative charge in the conjugate base (inductive effect). PH 4 + NH 4 + PH 4 + is more acidic because the larger P is more polarizable than the smaller N. A B C C N H H H H H H H H N C C H H H H H H H H H H A is more acidic because the N is sp 2 hybridized (compared to the 3 hybridized N in B .) A C C C N H H H H H H H H N C N H H H H H H H A is more acidic. The conjugate acid of C is strongly stabilized by resonance: N C N H H H H H H H N C N H H H H H H H There is resonance in the conjugate base, but it is better in the conjugate acid, making it less acidic.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Chem 10181, Fall 2007 EXAM 2 KEY October 16, 2007 2 (2)(a) (16 pts.) Cinnamic acid, (C 8 H 7 COOH, HCin) is derived from cinnamon; methylamine (CH 3 NH 2 ) is found in decomposing fish. 50 mL of a 0.840 M cinnamic acid solution is mixed with 50 mL of 0.500 M aqueous CH 3 2 . Compute the final concentrations of the species indicated in the box below, as well as the pH, after the solution comes to equilibrium. For HCin, K a = 3.65 × 10 -5 , pK a = 4.44; for CH 3 3 + , K a = 2.70 × 10 -11 , pK a = 10.57. [HCin] = 0.170 M [Cin ] = 0.250 M [CH 3 2 ] = 2.72 × 10 -7 M [CH 3 3 + ] = 0.250 M [H 3 O + ] = 2.48 x 10 -5 M pH = 4.61 Strongest acid = HCin, strongest base = CH 3 2 : [Don't forget dilution for the initial concs.!] Init.: 0.420 M 0.250 0 0 K a (HCin) HCin + CH 3 2 Cin - + CH 3 3 + K eq = ––––––––– Final: 0.170 M ~0 0.250 M 0.250 M K a (CH 3 3 + ) 3.65 x 10 = –––––––––– = 1.35 x 10 6 2.70 x 10 -11 [Cin - ][CH 3 3 + ] [0.250][0.250] ––––––––––––– = –––––––––––––– = 1.35 x 10 6 , so [CH 3 2 ] = [HCin][CH 3 2 ] [0.170][CH 3 2 ] 2.72 x 10 -7 M To find [H 3 O + ], use another acid-base equilibrium, such as: [H 3 O + ][Cin - ] 3 O + ][0.250] HCin + H 2 O Cin - + H 3 O + K = –––––––––– = ––––––––––––– = 3.65x10 [HCin] [0.170] 3 O + ] = 2.48 x 10 pH = -log 10 3 O + ] = 4.61
Background image of page 2
Chem 10181, Fall 2007 EXAM 2 KEY October 16, 2007 3 (b) (10 pts.) If gaseous methylamine is bubbled through a solution of cinnamic acid in an
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/17/2008 for the course CHEM 10181 taught by Professor Brown during the Fall '08 term at Notre Dame.

Page1 / 7

Exam 2 Key - Chem 10181, Fall 2007 (1) EXAM 2 KEY October...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online