++++Chem 162-2009 Chapter 16-Equilibria practice problems

++++Chem 162-2009 Chapter 16-Equilibria practice problems -...

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CHAPTER 16 EQUILIBRIA Problems to prepare students for Chem 162 hourly exam III. “Equilibria” covers Zumdahl 2 nd half of chapter 15. SOLUBILITY PRODUCT COMPLEX ION EQUILIBRIA E. Tavss 4/06 Chapter 15B practice problems 1
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SOLUBILITY PRODUCT 10 Chem 162-2008 Final Exam + Answers Chapter 16 – Slightly Soluble Salts and Complex Ion Equilibria Solubility product The solubility product, K sp , of PbCl 2 is 1.6 × 10 -5 . What mass of KCl(s) must be dissolved in 150.00 mL of 1.15M Pb(NO 3 ) 2 to produce a solution that is just saturated in PbCl 2 ? A . 0.0417 g B. 1.56×10 -5 g C. 0.278 g D. 0.00373 g E. 9.28×10 -5 g PbCl 2 (s) Pb 2+ + 2Cl - PbCl 2 (s) Pb 2+ + 2Cl - Initial Y 0 0 Change -X Equilibrium 1.15M X [Pb 2+ ] [Cl - ] 2 = K sp 1.15 x X 2 = 1.6 x 10 -5 X = 3.73 x 10 -3 M Cl - = 3.73 x 10 -3 M KCl 3.73 x 10 -3 mol/L x 0.150L = 5.595 x 10 -4 mol KCl in 150.00 mL 5.595 x 10 -4 mol KCl x 74.55 g/mol = 0.0417 g KCl A Chapter 15B practice problems 2
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8 Chem 162-2008 Final Exam + Answers Chapter 16 – Slightly Soluble Salts and Complex Ion Equilibria Solubility product A solution is 0.0100M in both CrO 4 2- and SO 4 2- . Slowly, Pb(NO 3 ) 2 is added to this solution. What is the [CrO 4 2- ] that remains in solution when PbSO 4 first begins to precipitate? K sp of PbCrO 4 = 2.8 × 10 -13 ; K sp of PbSO 4 = 1.6 × 10 -8 . A. 2.8×10 -13 B. 1.60×10 -6 C. 1.60×10 -8 D. 1.75×10 -5 E . 1.75×10 -7 PbCrO 4 Pb 2+ + CrO 4 2- [Pb 2+ ][CrO 4 2- ] = 2.8x10 -13 [Pb 2+ ][0.0100] = 2.8x10 -13 [Pb 2+ ] = 2.8x10 -11 M for PbCrO 4 to just begin to precipitate. PbSO 4 Pb 2+ + SO 4 2- [Pb 2+ ][SO 4 2- ] = 1.6x10 -8 [Pb 2+ ][0.0100] = 1.6x10 -8 [Pb 2+ ] = 1.6x10 -6 M for PbSO 4 to just begin to precipitate. Hence, the Pb 2+ begins to precipitate the CrO 4 2- . But when the concentration of Pb 2+ from the burette just reaches 1.6 x 10 -6 M the reaction is stopped so that no PbSO 4 will begin precipitate. At that concentration of Pb 2+ the concentration of the remaining CrO 4 2- in the solution can be calculated: [1.6x10 -6 ][CrO 4 2- ] = 2.8x10 -13 [CrO 4 2- ] = 1.75x10 -7 M remains in solution, a tiny amount compared to the 0.01M [CrO 4 2- ] at the beginning. E Chapter 15B practice problems 3
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7 Chem 162-2008 Final Exam + Answers Chapter 16 – Slightly Soluble Salts and Complex Ion Equilibria Solubility product The addition of which compound will increase the solubility of ZnCO 3 ? A. Zn(NO 3 ) 2 B. Na 2 CO 3 C. NaCl D. NaOH E . HCl ZnS Zn 2+ + S 2- A. No. Zn(NO 3 ) 2 provides Zn 2+ which, by the common ion effect, will force the equilibrium equation to the left, making ZnCO 3 less soluble. B. No. Na 2 CO 3 provides CO 3 2- which, by the common ion effect, will force the equilibrium equation to the left, making ZnCO 3 less soluble. C.
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This note was uploaded on 03/29/2011 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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++++Chem 162-2009 Chapter 16-Equilibria practice problems -...

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