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# solution_pdf5 - rahman(pr5989 Homework 5 owers(53880 This...

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rahman (pr5989) – Homework 5 – flowers – (53880) 1 This print-out should have 37 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Homework 5. Due 21 October 2008 at 7PM 001 10.0 points Solutions that contain a weak acid plus a salt of the weak acid are always more acidic than solutions that contain the same concentration of the weak acid alone. 1. False correct 2. True Explanation: HA + H 2 O H 3 O + + A This is a buffer system. The salt of the weak acid HA dissociates in solution, producing additional anions (A ) of the acid, which remove H 3 O + ions from so- lution and send the equilibrium to the left, according to Le Chˆ atelier s principle. Re- moval of H 3 O + from the system renders the solution less acidic. 002 10.0 points A buffer (pH 3.70) was prepared by mixing 1.00 mole of formic acid and 1.00 mole of sodium formate to form an aqueous solution with a total volume of 1.00 L. To 400 mL of this solution was added 50.0 mL of 1.00 M NaOH. What is the pH of this solution? 1. 4.52 2. 3.81 correct 3. 3.63 4. 4.25 5. 4.39 Explanation: [HF] = 1 M [NaOH] = 1 M [F ] = 1 M pH ini = 3.70 Initial condition (ini): n HF = 400 × 1 . 0 = 400 mmol n NaOH = 50 . 0 × 1 . 0 = 50 mmol n Na + = 400 × 1 . 0 = 400 mmol n F - = 400 × 1 . 0 = 400 mmol HF + NaOH Na + + F + H 2 O ini 400 50 . 0 400 400 Δ - 50 - 50 50 50 fin 350 0 450 450 Na + is a spectator ion. HF / F is a buffer system. Since [HF] = [F ] in the original buffer p K a = pH ini = 3 . 70, and pH fin = p K a + log parenleftBigg bracketleftbig F bracketrightbig [HF] parenrightBigg = 3 . 70 + log parenleftbigg 450 350 parenrightbigg = 3 . 80914 003 (part 1 of 2) 10.0 points Below is the pH curve of a weak acid (HA) titrated with strong base. 0 5 10 15 20 25 30 35 40 45 50 55 60 0 1 2 3 4 5 6 7 8 9 10 11 12 13 Titration Curve Amount of Base added (mL) Acid/Base concentration (pH) What is the pH at the equivalence point of this titration? Correct answer: 6 . 2 pH. Explanation:

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rahman (pr5989) – Homework 5 – flowers – (53880) 2 The inflection points are shown below. 0 5 10 15 20 25 30 35 40 45 50 55 60 0 1 2 3 4 5 6 7 8 9 10 11 12 13 Titration Curve Amount of Base added (mL) Acid/Base concentration (pH) (18, 3) (36, 6 . 2) 004 (part 2 of 2) 10.0 points How much base much be added to make the solution equalized? Correct answer: 36 mL. Explanation: 005 (part 1 of 2) 10.0 points Suppose that two hydroxides, MOH and M (OH) 2 , both have a K sp of 5 . 83 × 10 12 and that initially both cations are present in a solution at concentrations of 0 . 001 mol/L. At what pH will the hydroxide precipitate when solid NaOH is added to the solution? Correct answer: 5 . 76567. Explanation: K sp = 5 . 83 × 10 12 [M 2+ ] = [M + ] = 0 . 001 mol/L For MOH: MOH(s) M + (aq) + OH (aq) K sp = [M + ] [OH ] [OH ] = K sp [M + ] = 5 . 83 × 10 12 0 . 001 = 5 . 83 × 10 9 pOH = - log[OH ] = 8 . 23433 pH = 14 - pOH = 5 . 76567 For M (OH) 2 : M (OH) 2 (s) M 2+ (aq) + 2 OH (aq) K sp = [M + ] [OH ] 2 [OH ] 2 = K sp [M + ] = 5 . 83 × 10 12 0 . 001 = 5 . 83 × 10 9 [OH ] = radicalbig 5 . 83 × 10 9 = 7 . 63544 × 10 5 pOH = - log[OH ] = 4 . 11717 pH = 14 - pOH = 9 . 88283 MOH will precipitate first at pH = 5 . 76567.
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solution_pdf5 - rahman(pr5989 Homework 5 owers(53880 This...

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