Spring 09 Recitation Quiz 3

Spring 09 Recitation Quiz 3 - CHEM 0120 Golde SPRING 2009...

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CHEM 0120 SPRING 2009 Monday 1:00 PM Golde QUIZ 3 WEEK 4 January 26, 2009 Name_______________________________ For ax 2 + bx + c = 0, x = - ± - b b ac a 2 4 2 ; K w = 1.00 x 10 -14 1. (Critical Ideas) Give the formula of one cation which has no acid-base properties and the formula of one anion, which has no acid-base properties. 2. The ionization of the weak acid, chlorous acid, HClO 2 can be written: HClO 2 (aq) + H 2 O(l) H 3 O + (aq) + ClO 2 - (aq), K c = 2.70 x 10 -5 A solution is prepared, which initially contains 0.0410M HClO 2 and no H 3 O + ions. a) Show that [H 3 O + ] = 1.04 x 10 -3 M, when the reaction comes to equilibrium . b) Determine pH and pOH of the solution at equilibrium. ANSWERS 1. Cation: Na + , Ca 2+ , etc. Anion: Cl - , Br - , I - , NO 3 - , ClO 4 - , SO 4 2- 2. a) HClO 2 + H 2 O ClO 2 - + H 3 O + I 0.0410 0 0 C –x +x +x E 0.0410 - x x x 2 x 0.0410 x - = 2.70 x 10 -5 . Apply the approximation 0.0410 - x = 0.0410 x 2 = 2.70 x 10 -5 x 0.0410 = 1.107 x 10 -6 x = [H 3 O + ] = 1.05 x 10 -3 M (exact method yields x = 1.04 x 10 -3 ) [If student correctly substitutes the given value of [H 3 O + ] in the equilibrium constant equation, give full credit.] b) pH = -log(1.04 x 10 -3 ) = 2.98; pOH = 14.00 – 2.98 = 11.02 CHEM 0120 SPRING 2009 Monday 6:00 PM Golde QUIZ 3 WEEK 4 January 26, 2009 For Make-up Only Regular recitation Day: Time: instructor: Quiz Score: Group score: Scores recorded by:
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Name_______________________________ K w = 1.00 x 10 -14 1. Write the equation for the ionization of the weak acid benzoic acid, C 6 H 5 COOH, in water. Be sure to show the correct charge on each ion. 2. N 2 H 4 is a weak base and ionizes as follows: N 2 H 4 (aq) + H 2 O(l) OH - (aq) + N 2 H 5 + (aq), K c = ? A solution of N 2 H 4 had an initial concentration (of N 2 H 4 ) of 0.0855M. No OH - or N 2 H 5 + was present initially. At equilibrium after the above ionization had taken place, it was found that the pH of the solution was 10.60. a) Using the given pH value, determine the OH - concentration at equilibrium. b) Set up an I-C-E table which shows all the initial, change and equilibrium concentrations for the reaction. c) Then determine the K c of this reaction. ANSWERS 1. C 6 H 5 COOH(aq) + H 2 O(l) C 6 H 5 COO - (aq) + H 3 O + (aq) 2. a) pOH = 14.00 – 10.60 = 3.40 [OH - ] = inv log(-3.40) = 3.98 x 10 -4 M (b) N 2 H 4 (aq) + H 2 O(l) OH - (aq) + N 2 H 5 + (aq), K c = ? Initial (M)
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Spring 09 Recitation Quiz 3 - CHEM 0120 Golde SPRING 2009...

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