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Unformatted text preview: Version 135 – 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 Bromcresol green is an acid base indicator and changes from yellow to blue in the pH range of 3.8 to 5.4. Which of the following is the best estimate of the p K a of the indicator? 1. 9.2 2. 7.0 3. 5.4 4. 4.6 correct 5. 3.8 6. It is impossible to even guess without more information. 7. 2.1 Explanation: The p K a for an indicator is usually centered in the middle of its pH range. The center of this range is 4.6. 002 10.0 points The volume of 0.100 M NaOH required for the titration of 100 mL of 0.100 M CH 3 COOH to the equivalence point would be (larger than, the same as, smaller than) if the acid were 100 mL of 0.100 M HCl. The equivalence point for the HCl titration would come at (a higher, a lower, the same) pH than for the CH 3 COOH titration. 1. larger than; a lower 2. smaller than; the same 3. larger than; a higher 4. the same as; a higher 5. the same; a lower correct Explanation: [NaOH] = 0.1 M [CH 3 COOH] = 0.1 M [HCl] = 0.1 M V HCl = 100 mL V CH 3 COOH = 100 mL The amount of acid required to react a given number of moles of base depends on the total number of moles of H 3 O + the acid can provide. 100 mL of 0.1 M CH 3 COOH and 100 mL of 0.1 M HCl both provide 10.0 mmol H 3 O + . However, because CH 3 COOH is a weak acid and HCl is a strong acid the pH at the equivalence points will differ. In the case of HCl the spectator ions present do not hydrolyze and the endpoint pH is 7. In the case of CH 3 COOH, the CH 3 COO − tritra- tion product hydrolyzes, producing OH − ions which raise the pH above 7. 003 10.0 points A metal chloride (MCl 3 ) has a K sp equal to 3 . 4 × 10 − 9 . What is the concentration of Cl − in a saturated solution of MCl 3 ? 1. 10 ppm 2. 469 ppm 3. 812 ppm 4. 119 ppm 5. 2520 ppm 6. 238 ppm 7. 356 ppm correct Explanation: Ksp = (x)(3x) 3 x = (Ksp/27) . 25 x = 0 . 00334988 mol/L [Cl-] = 3x = 0 . 0100496 mol/L . 0100496 mol/L x 35.453 g/mol * 1000 mg/1 g = 356 . 29 ppm 004 10.0 points Sn 2+ is added to a solution of 0 . 00135 M X − and 0 . 0019 M Z 2 − . The SnZ precipitates first ( K sp = 1 × 10 − 12 ), followed by the SnX 2 Version 135 – Exam 3 – Mccord – (52450) 2 ( K sp = 1 . 8 × 10 − 14 ). What percentage of the Z 2 − is still in solution (not precipitated) when the SnX 2 just starts to precipitate? 1. 9.75806 2. 30.1932 3. 27.7778 4. 5.32895 5. 13.5813 6. 16.5681 7. 6.31381 8. 12.3671 9. 14.7947 10. 7.8125 Correct answer: 5 . 32895%. Explanation: [X − ] = 0 . 00135 M K sp SnZ = 1 × 10 − 12 [Z 2 − ] = 0 . 0019 M K sp SnX 2 = 1 . 8 × 10 − 14 The two relevant equilibria are SnX 2 (s) ⇀ ↽ Sn 2+ + 2 X − K sp = [Sn 2+ ][X − ] 2 SnZ(s) ⇀ ↽ Sn 2+ + Z 2 − K sp = [Sn 2+ ] [Z 2 − ] SnX 2 would precipitate when K sp = [Sn 2+ ] [X − ] 2 [Sn 2+ ] = K sp [ X − ] 2 = 1 . 8 ×...
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This note was uploaded on 06/15/2010 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.

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exam3_pdf - Version 135 – Exam 3 – Mccord –(52450 1...

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