Quiz-2-07F-Solutions

Quiz-2-07F-Solutions - 1 CHEM 119 H Quiz 2 11/06/07 USE...

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1 CHEM 119 H Quiz 2 11/06/07 ____________________________________ USE PENCIL! ERASE CLEANLY! If we cannot I have neither given nor received aid while follow your work quickly, it is wrong. taking this quiz. 1. (50) A weak acid, HZ, is dissolved in water. pK A {HZ} = 2.810 ε {HZ, 450 nm} = 2348 L/cm*mol ε {Z - , 450 nm} = 0 a. (20) Calculate the absorbance, A, of 0.00167 F{HZ} in a 1.15 cm cell. A = 1.782 [] + 3 -+ + A 33 2 + 3 -2.810 + 3 + 3 Only HZ absorbs, but some HZ dissociates and [HZ] F{HZ} HO Z K = and standard approximation gives [Z ] = [H O ] and [HZ] = F- [H O ]; HZ Therefore, 10 = H O .00167- H O ⎡⎤ ⎣⎦ ⇒= - -4 [Z ]=0.00101M and [HZ] = (0.00167 - 0.00101) = 6.6*10 M () 2348L 0.00066mol A = 1.15cm = 1.782 cm*mol L ⎞⎞ ⎛⎛ ⎜⎟ ⎜ ⎝⎝ ⎠⎠ b. (20) Calculate the % Transmittance, in a 11.5 cm cell, of the solution prepared when 1.00 mL of 0.00167 F{HZ} is diluted to a final volume of 10.0 mL. %T = 39.4 % 1.00 mL Dilution gives a new formality: F(diluted) = F{Original} 0.000167M 10.0 mL = ⎜⎟ If the extent of ionization of the acid did not change, the absorbance and % T would be the same because A a = 2348*1.15*C a and A b = 2348*(11.5)(C a /10). Hence, % T = 100*10 -1.782 = 1.7% (1.65 %). BUT the extent of dissociation does depend on concentration and the extent of dissociation increases with decreasing concentration (Le Chatelier, again). Hence 2 2.81 -5 [Z ] 10 [Z ] 0.000152 M & [HZ] = 0.000167-0.000152 = 1.5*10 M 0.000167 [Z ] −− =⇒ = A = 2348*11.5*1.5*10 -5 = 0.405 %T = 10 -0.405 = 39.4% c. (10) Calculate the absorbance of 0.00167 F{HZ} in a 1.15 cm cell, if
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This note was uploaded on 04/13/2008 for the course CHEM 119H taught by Professor Munson during the Fall '07 term at University of Delaware.

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Quiz-2-07F-Solutions - 1 CHEM 119 H Quiz 2 11/06/07 USE...

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