Fahkreddine_HW6

Fahkreddine_HW6 - tu (pyt67) – Homework 6 – Fakhreddine...

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Unformatted text preview: tu (pyt67) – Homework 6 – Fakhreddine – (52425) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 1.0 points Some solid NH 4 Cl is added to a water solution of NH 3 . Which of the following statements is correct? 1. pH of the solution increases. 2. [H 3 O + ] of the solution remains the same. 3. pH of the solution decreases. correct 4. [OH- ] of the solution increases. Explanation: 002 1.0 points Some solid KClO 2 is added to a water solution of HClO 2 . Which of the following statements is correct? 1. [OH- ] of the solution decreases. 2. [H 3 O + ] will remain the same. 3. pH of the solution increases. correct 4. pH of the solution decreases. Explanation: 003 1.0 points Calculate the pH of a solution containing 0.10 M CH 3 COOH and 0.10 M NaCH 3 COO. 1. 7.12 2. 8.27 3. 2.89 4. 5.78 5. 4.74 correct Explanation: [CH 3 COOH] = 0 . 10 M K a = 1 . 8 × 10- 5 [CH 3 COO- ] = 0 . 10 M This is an acetate buffer in which [CH 3 COOH] = [CH 3 COO- ], so pH = p K a =- log(1 . 8 × 10- 5 ) = 4 . 74473 004 1.0 points What is the approximate pH of a solution in which the concentration of benzoic acid is 1.0 M and the concentration of sodium benzoate is 0.10 M? 1. 3.20 correct 2. 2.64 3. 5.20 4. 6.64 5. 1.65 Explanation: bracketleftbig C 6 H 5 COO- bracketrightbig = 1 . 0 M [C 6 H 5 COOH] = 0 . 1 M pH = p K a + log parenleftBigg bracketleftbig C 6 H 5 COO- bracketrightbig [C 6 H 5 COOH] parenrightBigg =- log ( 6 . 3 × 10- 5 ) + log . 10 1 . = 3 . 20066 005 1.0 points What is the pH of a solution which is 0.600 M in dimethylamine ((CH 3 ) 2 NH) and 0.400 M in dimethylamine hydrochloride ( (CH 3 ) 2 NH + 2 Cl- ) ? K b for dimethylamine = 7 . 4 × 10- 4 . 1. 10.78 2. 10.87 3. 10.69 4. 11.21 tu (pyt67) – Homework 6 – Fakhreddine – (52425) 2 5. 11.05 correct 6. 3.31 7. 2.95 Explanation: K w = 1 × 10- 14 K a = 0 . 00074 [(CH 3 ) 2 NH] = 0.6 M [(CH 3 ) 2 NH + 2 ] = 0.4 M K a , (CH 3 ) 2 NH + 2 = K w K b , (CH 3 ) 2 NH Applying the Henderson-Hasselbalch equa- tion, pH = p K a + log parenleftbigg [(CH 3 ) 2 NH] [(CH 3 ) 2 NH + 2 ] parenrightbigg =- log parenleftbigg K w K a parenrightbigg + log parenleftbigg [(CH 3 ) 2 NH] [(CH 3 ) 2 NH + 2 ] parenrightbigg =- log parenleftbigg 1 × 10- 14 . 00074 parenrightbigg + log parenleftbigg . 6 . 4 parenrightbigg = 11 . 0453 006 1.0 points A 0.50 M solution of dimethylamine (a weak base with K b = 7 . 4 × 10- 4 ) would be resistant to large changes of pH Z1) when a strong acid is added, but not if a strong base is added. Z2) when a strong base is added, but not if a strong acid is added. Z3) when either a strong acid or a strong base is added....
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This note was uploaded on 03/03/2012 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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Fahkreddine_HW6 - tu (pyt67) – Homework 6 – Fakhreddine...

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