solutionEXAM3_pdf

solutionEXAM3_pdf - Version 075 – Exam 3 – Fakhreddine...

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Unformatted text preview: Version 075 – Exam 3 – Fakhreddine – (52395) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points 50 mL of 2 . 0 M acetic acid (CH 3 COOH) was titrated with 1 . 0 M NaOH. What is the pH at the equivalence point of this titration? The ionization constant is 1 . 8 × 10- 5 . 1. 9.52 2. 9.20 3. 10.52 4. 7.00 5. 9.28 correct Explanation: [CH 3 COOH] = 2.0 M [NaOH] = 1.0 M V CH 3 COOH = 50 mL K a = 1 . 8 × 10- 5 Initially, [CH 3 COOH] = (50 mL)(2.0 M) = 100 mmol CH 3 COOH+NaOH → Na + +CH 3 COO- + H 2 O ini 100 100 Δ − 100 − 100 100 100 fin 100 100 Na + is a spectator ion. The volume of NaOH required is V NaOH = 100 mmol × 1 . 0 mL 1 . 0 mmol = 100 mL Total volume = 150 mL. [CH 3 COO- ] = 100 mmol 150 mL = 0 . 666667 M Hydrolysis: CH 3 COO- +H 2 O ⇀ ↽ CH 3 COOH + OH- ini, M . 666667 Δ, M − x x x eq, M 0 . 666667 − x x x K b = K w K a CH 3 COOH = [CH 3 COOH][OH- ] [CH 3 COO- ] 1 × 10- 14 1 . 8 × 10- 5 = x 2 . 666667 − x Assume x ≪ . 666667 x = [OH- ] = 1 . 9245 × 10- 5 M ( ≪ . 666667) pOH = − log(1 . 9245 × 10- 5 M) = 4 . 71568 pH = 14 − 4 . 71568 = 9 . 28432 002 10.0 points What volume of 0.585 M Ca(OH) 2 would be needed to neutralize 15.8 L of 1.51 M HCl? 1. 3.06 L 2. 6.12 L 3. 20.4 L correct 4. 40.8 L 5. 12.2 L Explanation: [Ca(OH) 2 ] = 0.585 M V HCl = 15.8 L [HCl] = 1.51 M The balanced equation for this neutraliza- tion reaction is 2 HCl + Ca(OH) 2 → CaCl 2 + 2 H 2 O We determine the moles of HCl present: ? mol HCl = 15 . 8 L soln × 1 . 51 mol HCl 1 L soln = 23 . 86 mol HCl Using the mole ratio from the chemical equation we calculate the moles of Ca(OH) 2 needed to react with this amount of HCl: ? mol Ca(OH) 2 = 23 . 86 mol HCl × 1 mol Ca(OH) 2 2 mol HCl = 11 . 93 mol Ca(OH) 2 We use the molarity of the Ca(OH) 2 so- lution to convert from moles to volume of Ca(OH) 2 : ? L Ca(OH) 2 = 11 . 93 mol Ca(OH) 2 × 1 L soln . 585 mol Ca(OH) 2 = 20 . 4 L Ca(OH) 2 Version 075 – Exam 3 – Fakhreddine – (52395) 2 003 10.0 points Which of the following compounds has the LOWEST molar solubility in water? 1. Al(OH) 3 , K sp = 1 . 9 × 10- 33 2. PbSO 4 , K sp = 1 . 8 × 10- 8 3. Sn(OH) 2 , K sp = 2 . × 10- 27 correct 4. AuCl 3 , K sp = 3 . 6 × 10- 29 Explanation: 004 10.0 points Assume the molar solubility of aluminum carbonate (Al 2 (CO 3 ) 3 ) is represented as x . Which of the following expressions correctly expresses the relationship between the molar solubility of aluminum carbonate and the sol- ubility product constant ( K sp ) for this com- pound? 1. K sp = 36 x 5 2. K sp = 54 x 5 3. K sp = 5 x 2 4. K sp = 6 x 5 5. K sp = 27 x 4 6. K sp = 108 x 6 7. K sp = 12 x 3 8. K sp = 108 x 5 correct Explanation: 005 10.0 points What is the pH of the solution resulting from the addition of 40.0 mL of 0.200 M HClO 4 to 20.0 mL of 0.150 M NaOH?...
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solutionEXAM3_pdf - Version 075 – Exam 3 – Fakhreddine...

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