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CH 302 - [McCord] [Spring 2010] - Test 3

# CH 302 - [McCord] [Spring 2010] - Test 3 - Version 269 Exam...

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Version 269 – Exam 3 – Mccord – (52450) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider the following electrode reactions: Fe 3+ + 1 e Fe 2+ E 0 = +0 . 771 V I 2 + 2 e 2 I E 0 = 0 . 535 V What would be E 0 cell for the spontaneous reaction? 1. 1.306 V 2. +0.236 V correct 3. 0.236 V 4. +1.007 V 5. +1.306 V Explanation: The anode is the place where oxidation oc- curs, while the cathode is the place where reduction occurs. So looking at the overall re- action, Fe 3+ is reduced to form Fe 2+ , because it gains electrons. So Fe 3+ | Fe 2+ is the reac- tion at the cathode. 2 I has to lose 2 e to form I 2 , so it has to be oxidized and this has to occur at the anode. E 0 net = E 0 cathode E 0 anode = +0 . 771 (+0 . 535) = +0 . 236 V 002 10.0 points The equivalence point during titration of an acid with a base may not occur at a pH of 7.0 because 1. hydrolysis of the salt produced may make the solution acidic or basic. correct 2. A solution containing a weak acid or a weak base could never be neutral. 3. The statement is false; the pH must be 7.0 at the equivalence point. 4. the indicator may be one that changes color at some other pH. 5. if a strong acid or base is used, it is com- pletely ionized and therefore not neutral. Explanation: If a weak acid or a weak base is involved in the titration its conjugate which forms at the equivalence point hydrolyzes to produce excess H 3 O + in the case of a weak base or excess OH in the case of a weak acid. weak acid: HA + OH H 2 O + A A + H 2 O HA + OH or weak base: B + H 3 O + H 2 O + BH + BH + + H 2 O B + H 3 O + 003 10.0 points Choose the effective pH range of a HF NaF buffer. For HF, K a = 3 . 5 × 10 4 . 1. 0.7 to 2.7 2. 2.5 to 4.5 correct 3. 5.0 to 7.0 4. 9.6 to 11.6 5. 6.0 to 8.0 Explanation: 004 10.0 points Calculate the solubility of iron(III) hydrox- ide at pH = 4 . 5. The solubility product of iron(III) hydroxide is 2 × 10 39 . 1. 3.99052e-24 2. 3.99052e-15 3. 5.02377e-07 4. 6.32456e-11 5. 5.02377e-28 6. 3.16979e-20 7. 2.0e-06

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Version 269 – Exam 3 – Mccord – (52450) 2 8. 3.16979e-23 9. 3.16979e-08 10. 1.00237e-27 Correct answer: 6 . 32456 × 10 11 . Explanation: K sp = 2 × 10 39 pH = 4 . 5 Let S = molar solubility pOH = 14 pH = 14 4 . 5 = 9 . 5 [OH ] = 10 pOH = 3 . 16228 × 10 10 The reaction is Fe 3+ (aq) + 3 OH (aq) Fe(OH) 3 (s) K sp = [Fe 3+ ] [OH ] 3 S = K sp [OH ] 3 = 2 × 10 39 (3 . 16228 × 10 10 ) 3 = 6 . 32456 × 10 11 005 10.0 points Which of the following pairs of solutions would result in a buffer upon mixing? 1. 3 L of 0 . 3 M ammonia; 2 L of 0 . 4 M HI correct 2. 3 . 2 L of 0 . 6 M NaCl; 4 . 3 L of 0 . 3 M KClO 4 3. 3 L of 0 . 2 M formic acid; 1 L of 0 . 6 M NaOH 4. 2 L of 0 . 2 M HClO 4 ; 3 L of 0 . 5 M CsOH Explanation: A buffer prepared by a neutralization re- action requires a weak acid mixed with less strong base or a weak base mixed with less strong acid. The only pair of solutions which satisfies this constraint is 3 L of 0 . 3 M ammo- nia; 2 L of 0 . 4 M HI.
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CH 302 - [McCord] [Spring 2010] - Test 3 - Version 269 Exam...

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