Chem 162-2011 exam II draft

Chem 162-2011 exam II draft - Chemistry 162 Exam II March...

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Chemistry 162 Exam II March 23, 2011 Chem 162-2011 Hourly Exam II + Answers Acid and base equilibria calculations 1. At 5 o C, the pH of 0.10 M NaOH = 13.74. What is the value of K w at 5 o C? A. 1.0 x 10 -15 B . 1.8 x 10 -15 C. 5.6 x 10 -15 D. 2.8 x 10 -14 E. 5.6 x 10 -14 pH = 13.74 [H + ] = 10 -pH = 10 -13.74 = 1.82 x 10 -14 [OH - ] = 0.10M K w = [H + ][OH - ] = [1.82 x 10 -14 ][0.10] = 1.82 x 10 -15 This makes sense given the facts that K w for water is 1 x 10 -14 at 25 o C, and that autoionization of water is an endothermic reaction . H 2 O + H 2 O H 3 O + + OH - K w at 25 o C = 1 x 10 -14 H = positive Hence, according to LeChatelier, removal of heat (i.e., lowering the temperature) shifts the equilibrium to the left, correspondingly decreasing the equilibrium constant. Note: 1.8 x 10 -15 is a smaller K than 1.0 x 10 -14 . Chem 162-2011 Hourly Exam II + Answers Chapter 16 – Equilibria Solubility product 2. In a saturated solution of silver phosphate, Ag 3 PO 4 , [Ag + ] = 5.3 x 10 -5 M. Calculate K sp for Ag 3 PO 4 . A. 7.9 x 10 -18 B. 6.0 x 10 -13 C . 2.6 x 10 -18 D. 6.0 x 10 -17 E. 5.2 x 10 -19 Ag 3 PO 4 (solid) 3Ag + (aq) + PO 4 3- (aq) Ag 3 PO 4 (solid) 3Ag + (aq) + PO 4 3- (aq) Initial Y 0 0 Change -X +3X +X Equilibrium Y-X 5.3 x 10 -5 +X 0 + 3X = 5.3 x 10 -5 X = 1.77 x 10 -5 1
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[Ag + ] 3 [PO 4 3- ] = K sp [5.3x10 -5 ] 3 [1.77x10 -5 ] = K sp K sp = 2.64 x 10 -18 Chem 162-2011 Hourly Exam II + Answers Chapter 16 – Equilibria Solubility product 3. PbI 2 (s) Pb 2 + (aq) + 2I - (aq) K sp = 7.1 x 10 -9 What amount of PbI 2 (s) can dissolve in 2.0 L of aqueous solution, which contains 0.50 mol of KI dissolved in it. A . 2.3 x 10 -7 mol B. 5.7 x 10 -8 mol C. 2.8 x 10 -8 mol D. 1.8 x 10 -9 mol E. 4.6 x 10 -7 mol Solve the problem based on 1.0L; then recalculate for 2.0L. PbI 2 (s) Pb 2 + (aq) + 2I - (aq) K sp = 7.1 x 10 -9 PbI 2 (s) Pb 2 + (aq) + 2I - (aq) Initial Y 0 0.50mol/2L= 0.25M Change -X +X +2X Equilibrium Y-X +X 0.25+2X ([Pb 2+ ][I - ] 2 ) = K sp ([X][0.25+2X] 2 ) = 7.1x10 -9 Quadratic equation, so use small K rule. ([X][0.25] 2 ) = 7.1x10 -9 X = 1.136 x 10 -7 M Solubility in 2 liters = 2.27 x 10 -7 mol PbI 2 dissolved. Chem 162-2011 Hourly Exam II + Answers Buffer calculations 4. NH 3 is a weak base with K b = 1.8 x 10 -5 . A buffer is prepared by dissolving 0.80 moles of NH 3 and 0.80 moles of NH 4 Cl in 1.00 L of aqueous solution. If 0.10 mol of NaOH is added to 250 mL of this buffer, what is the pH of the resultant solution? A . 9.73 B. 10.54 C. 5.22 D. 4.27 2
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E. 8.78 Adding 0.10 mol of NaOH to 0.250L is the same as adding 0.40 mol to 1.000L. NH 4 + + OH - → NH 3 + H 2 O Use large K rule. NH
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Chem 162-2011 exam II draft - Chemistry 162 Exam II March...

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