HW5solutions

# HW5solutions - Niu(qn269 – Homework 5 – flowers...

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Unformatted text preview: Niu (qn269) – Homework 5 – flowers – (53885) 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 7PM 001 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. 3.81 correct 2. 4.52 3. 4.39 4. 3.63 5. 4.25 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 . × 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 . 400 400 Δ- 50- 50 50 50 fin 350 450 450 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 002 10.0 points If 100 mL of 0.050 molar HCl is added to 50 mL of 0.100 molar NH 3 , the resulting solution will be 1. basic because excess NH 3 is present. 2. acidic because of hydrolysis of NH + 4 . cor- rect 3. basic because of hydrolysis of NH + 4 . 4. acidic because excess HCl is present. 5. neutral. Explanation: V HCl = 100 mL [HCl] = 0.05 M V NH 3 = 50 mL [NH 3 ] = 0.1 M n HCl = (100 mL)(0.05 M) = 5 mmol n NH 3 = (50 mL)(0.1 M) = 5 mmol NH 3 + HCl-→ NH + 4 + Cl − Equal numbers of moles of a strong acid and a weak base have reacted, so this is the equivalence point. The conjugate acid of the weak base hydrolyzes to produce H 3 O + and so the solution is acidic. NH 4 + H 2 O ⇀ ↽ NH 3 + H 3 O + 003 (part 1 of 2) 10.0 points Below is the pH curve of a weak acid (HA) titrated with strong base. Niu (qn269) – Homework 5 – flowers – (53885) 2 0 5 10 15 20 25 30 35 40 45 50 55 60 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: 7 pH. Explanation: The inflection points are shown below. 0 5 10 15 20 25 30 35 40 45 50 55 60 1 2 3 4 5 6 7 8 9 10 11 12 13 Titration Curve Amount of Base added (mL) Acid/Base concentration(pH) (14, 3 . 9) (28, 7) 004 (part 2 of 2) 10.0 points How much base much be added to make the solution equalized? Correct answer: 28 mL. Explanation: 005 10.0 points A weak acid is titrated by a strong base. Which of the following features about the titration curve is NOT true? 1. When 0 mL of strong base is added, the pH of the solution is < 7....
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## This note was uploaded on 02/04/2009 for the course CH 53885 taught by Professor Flowers during the Fall '08 term at University of Texas.

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HW5solutions - Niu(qn269 – Homework 5 – flowers...

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