EXAM2solutions

# EXAM2solutions - Version 012 Exam 2 owers(53885 This...

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Version 012 – Exam 2 – fowers – (53885) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. Exam 3. No Books, No Notes. You may bring and use a scientiFc calculator. PLEASE PLEASE PLEASE bubble in your name, UTEID, and EXAM VERSION num- ber properly. Thank you and good luck 001 10.0 points To simulate blood conditions, a phosphate bu²er system with a pH = 7 . 4 is desired. What mass o± Na 2 HPO 4 must be added to 1 . 75 L o± 0 . 32 M NaH 2 PO 4 (aq) to prepare such a bu²er? 1. 46.1723 2. 148.411 3. 109.934 4. 362.783 5. 153.908 6. 80.8016 7. 73.6559 8. 85.7486 9. 123.126 10. 206.126 Correct answer: 123 . 126 g. Explanation: [NaH 2 PO 4 ] = 0 . 32 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 ] Using Henderson-Hasselbalch equation, pH = p K a + log p [HPO 2 - 4 ] [H 2 PO - 4 ] P 7 . 4 = 7 . 21 + log p [HPO 2 - 4 ] [H 2 PO - 4 ] P log p [HPO 2 - 4 ] [H 2 PO - 4 ] P = 0 . 19 [HPO 2 - 4 ] [H 2 PO - 4 ] = 10 0 . 19 = 1 . 54882 [Na 2 HPO 4 ] = (1 . 54882)(0 . 32 mol / L) = 0 . 495621 mol / L m Na 2 HPO 4 = 0 . 495621 mol(1 . 75 L) = 0 . 867337 mol Na 2 HPO 4 Na 2 HPO 4 = 141 . 959 g/mol mass = (0 . 867337)(141 . 959) = 123 . 126 g Na 2 HPO 4 002 10.0 points What is K sp ±or Ag 3 PO 4 , i± its molar solubility is 2 . 7 × 10 - 6 mol / L? 1. 1 . 4 × 10 - 21 correct 2. 5 . 3 × 10 - 23 3. 5 . 3 × 10 - 16 4. 1 . 7 × 10 - 14 5. 2 . 0 × 10 - 17 6. 7 . 3 × 10 - 12 7. 4 . 8 × 10 - 22 Explanation: S = 2 . 7 × 10 - 6 mol / L The solubility equilibrium is Ag 3 PO 4 (s) 3 Ag + (aq) + PO 3 - 4 (aq) [Ag + ] = 3 S = 8 . 1 × 10 - 6 mol / L [PO 3 - 4 ] = S = 2 . 7 × 10 - 6 mol / L K sp = [Ag + ] 3 [PO 3 - 4 ] = ( 8 . 1 × 10 - 6 ) 3 (2 . 7 × 10 - 6 ) = 1 . 43489 × 10 - 21 003 10.0 points

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Version 012 – Exam 2 – fowers – (53885) 2 What is the solubility in mol/L oF man- ganese(II) sul±de (MnS) given that its K sp value is 1 . 7 × 10 - 13 ? 1. 3.87298e-07 2. 4.79583e-07 3. 4.89898e-07 4. 4.3589e-07 5. 5.2915e-07 6. 5.0e-07 7. 4.12311e-07 8. 5.09902e-07 9. 4.24264e-07 10. 4.58258e-07 Correct answer: 4 . 12311 × 10 - 7 mol / L. Explanation: K sp oF MnS = 1 . 7 × 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 - ] 1 . 7 × 10 - 13 = x 2 x = r 1 . 7 × 10 - 13 = 4 . 12311 × 10 - 7 mol / L = solubility oF MnS 004 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 base. 2. a weak acid. correct 3. a strong base. 4. a strong acid. 5. neither an acid nor a base. Explanation: 005 10.0 points Calculate the resulting pH iF 365 mL oF 2.88 M HNO 3 is mixed with 335 mL oF 1.10 M Ca(OH) 2 solution. 1. 2.36 2. 1.46 3. 7.20 4. 0.350 correct 5. 0.067 6. 0.460 Explanation: V HNO 3 = 365 mL [HNO 3 ] = 2.88 M V Ca(OH) 2 = 335 mL [Ca(OH) 2 ] = 1.10 M To determine the pH oF the ±nal mixture, we need to determine how much H or OH is leFt over aFter the reaction. Remember that For complete neutralization we need H and OH in equal molar amounts: H + + OH - H 2 O ²irst calculate how many moles oF H + and OH - we have: ? mol H + = 0 . 365 L × 2 . 88 mol HNO 3 1 L × 1 mol H + 1 mol HNO 3 = 1 . 05 mol H + ? mol OH - = 0 . 335 L × 1 . 10 mol Ca(OH) 2 1 L × 2 mol OH - 1 mol Ca(OH) 2 = 0 . 737 mol OH -
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EXAM2solutions - Version 012 Exam 2 owers(53885 This...

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