WS9_key - Chem 142 Summer 2011 Page 1 of 16 Name

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Unformatted text preview: Chem 142 Summer 2011 Page 1 of 16 Name ________________________________________ Section ________________ Worksheet 9: Buffers, Titrations, and Solubility 1. What is the pH and percent dissociation of the acid in each of the following solutions? What is the pH after 50.0 mL of 3.00 M NaOH is added to the solution? What is the pH after 80.0 mL of 4.00 M HCl is added to the original solution? Rank the buffers in order of increasing buffer capacity based on your calculations. a. 0.200 M HC 2 H 3 O 2 (K a = 1.8 x 10-5 ) in the presence of 0.500 M NaC 2 H 3 O 2 in a total volume of 1.00 L pH = pK a + log C 2 H 3 O 2 ! " # $ % HC 2 H 3 O 2 [ ] & ( ) * + = ! log 1.8 , 10 ! 5 ( ) + log 0.500 0.200 & ( ) * + pH = 4.74 + 0.3979 = 5.14 K a = 1.8 ! 10 " 5 = H + # $ % & C 2 H 3 O 2 " # $ % & HC 2 H 3 O 2 [ ] = x 0.500 ( ) 0.200 x = H + ! " # $ = 7.2 % 10 & 6 M % dissoc . = 7.2 ! 10 " 6 0.200 ! 100% = 3.6 ! 10 " 3 % Addition of 50.00 mL of 3.00 M NaOH HC 2 H 3 O 2 ( aq ) + OH- ( aq ) ! C 2 H 3 O 2- ( aq ) + H 2 O ( l ) HC 2 H 3 O 2 OH- C 2 H 3 O 2- Initial (mmol) 200. 150. 500. Final (mmol) 50. 0 650. HC 2 H 3 O 2 ( aq ) ! C 2 H 3 O 2- ( aq ) + H + ( aq ) [HC 2 H 3 O 2 ] [C 2 H 3 O 2- ] [H + ] Initial 0.04762 0.6190 0 Change -x +x +x Equilibrium 0.04762-x 0.6190+x x Assume x is small compared to initial concentrations K a = 1.8 ! 10 " 5 = H + # $ % & C 2 H 3 O 2 " # $ % & HC 2 H 3 O 2 [ ] = x 0.6190 ( ) 0.04762 x = H + ! " # $ = 1.4 % 10 & 6 M pH = log H + ( ) = log 1.4 10 6 ( ) = 5.86 Chem 142 Summer 2011 Page 2 of 16 Addition of 80.00 mL of 4.00 M HCl C 2 H 3 O 2- ( aq ) + H + ( aq ) ! HC 2 H 3 O 2 ( aq ) C 2 H 3 O 2- H + HC 2 H 3 O 2 Initial (mmol) 500. 320. 200. Final (mmol) 180. 0 520. HC 2 H 3 O 2 ( aq ) ! C 2 H 3 O 2- ( aq ) + H + ( aq ) [HC 2 H 3 O 2 ] [C 2 H 3 O 2- ] [H + ] Initial 0.482 0.167 0 Change -x +x +x Equilibrium 0.482-x 0.167+x x Assume x is small compared to initial concentrations K a = 1.8 ! 10 " 5 = H + # $ % & C 2 H 3 O 2 " # $ % & HC 2 H 3 O 2 [ ] = x 0.167 ( ) 0.482 x = H + ! " # $ = 5.2 % 10 & 5 M pH = ! log H + " # $ % ( ) = ! log 5.2 & 10 ! 5 ( ) = 4.28 b. 0.200 M HC 3 H 5 O 2 (K a = 1.3 x 10-5 ) in the presence of 0.400 M NaC 3 H 5 O 2 in a total volume of 1.00 L pH = pK a + log C 3 H 5 O 2 ! " # $ % HC 3 H 5 O 2 [ ] & ( ) * + = ! log 1.3 , 10 ! 5 ( ) + log 0.400 0.200 & ( ) * + pH = 4.89 + 0.30 = 5.19 K a = 1.3 ! 10 " 5 = H + # $ % & C 3 H 5 O 2 " # $ % & HC 3 H 5 O 2 [ ] = x 0.400 ( ) 0.200 x = H + ! " # $ = 6.5 % 10 & 6 M % dissoc . = 6.5 ! 10 " 6 0.200 ! 100% = 3.3 ! 10 " 3 % Chem 142 Summer 2011 Page 2 of 16 Addition of 80.00 mL of 4.00 M HCl C 2 H 3 O 2- ( aq ) + H + ( aq ) ! HC 2 H 3 O 2 ( aq ) C 2 H 3 O 2- H + HC 2 H 3 O 2 Initial (mmol) 500. 320. 200....
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This note was uploaded on 01/19/2012 for the course CHEM 142A taught by Professor Campbell during the Summer '11 term at University of Washington.

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WS9_key - Chem 142 Summer 2011 Page 1 of 16 Name

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