Exam 2 Solutions - Version 370 Exam 2 flowers (53885) 1...

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Unformatted text preview: Version 370 Exam 2 flowers (53885) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Exam 3. No Books, No Notes. You may bring and use a scientific calculator. PLEASE PLEASE PLEASE bubble in your name, UTEID, and EXAM VERSION num- ber properly. Thank you and good luck 001 10.0 points What is the solubility in mol/L of man- ganese(II) sulfide (MnS) given that its K sp value is 2 . 3 10- 13 ? 1. 4.69042e-07 2. 4.3589e-07 3. 5.2915e-07 4. 4.24264e-07 5. 5.0e-07 6. 4.89898e-07 7. 4.79583e-07 8. 4.47214e-07 9. 5.09902e-07 10. 4.12311e-07 Correct answer: 4 . 79583 10- 7 mol / L. Explanation: K sp of MnS = 2 . 3 10- 13 solubility of MnS in mol/L = ? MnS(s) Mn 2+ (aq) + S 2- (aq) [Mn 2+ ] = [S 2- ] = x K sp = [Mn 2+ ] [S 2- ] 2 . 3 10- 13 = x 2 x = radicalbig 2 . 3 10- 13 = 4 . 79583 10- 7 mol / L = solubility of MnS 002 10.0 points A sample is dissolved in water. One out of every 1000 molecules dissociate and donate a proton. This is 1. a weak acid. correct 2. a strong base. 3. neither an acid nor a base. 4. a strong acid. 5. a weak base. Explanation: 003 10.0 points A solution with a higher pH is more acidic. 1. False correct 2. True 3. The answer cannot be determined with- out additional information. Explanation: pH = log [H 3 O + ], so as [H 3 O] increases, pH decreases. 004 10.0 points To simulate blood conditions, a phosphate buffer system with a pH = 7 . 4 is desired. What mass of Na 2 HPO 4 must be added to . 25 L of 0 . 19 M NaH 2 PO 4 (aq) to prepare such a buffer? 1. 97.2917 2. 62.6624 3. 10.4437 4. 294.623 5. 373.776 6. 76.9539 7. 57.1657 8. 82.4506 9. 415.551 10. 85.7486 Correct answer: 10 . 4437 g. Explanation: [NaH 2 PO 4 ] = 0 . 19 mol/L MW Na 2 HPO 4 = 141 . 959 g/mol The equilibrium involved is H 2 PO- 4 (aq) + H 2 O( ) H 3 O + (aq) + HPO 2- 4 (aq) K a2 = [H 3 O + ] [HPO 2- 4 ] [H 2 PO- 4 ] Version 370 Exam 2 flowers (53885) 2 Using Henderson-Hasselbalch equation, pH = p K a + log parenleftBigg [HPO 2- 4 ] [H 2 PO- 4 ] parenrightBigg 7 . 4 = 7 . 21 + log parenleftBigg [HPO 2- 4 ] [H 2 PO- 4 ] parenrightBigg log parenleftBigg [HPO 2- 4 ] [H 2 PO- 4 ] parenrightBigg = 0 . 19 [HPO 2- 4 ] [H 2 PO- 4 ] = 10 . 19 = 1 . 54882 [Na 2 HPO 4 ] = (1 . 54882)(0 . 19 mol / L) = 0 . 294275 mol / L m Na 2 HPO 4 = 0 . 294275 mol(0 . 25 L) = 0 . 0735688 mol Na 2 HPO 4 Na 2 HPO 4 = 141 . 959 g/mol mass = (0 . 0735688)(141 . 959) = 10 . 4437 g Na 2 HPO 4 005 10.0 points What is the approximate pH of a solution in which the concentration of formic acid is 0.80 M and the concentration of sodium formate is 0.20 M?...
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Exam 2 Solutions - Version 370 Exam 2 flowers (53885) 1...

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