ASA-Ksp - -5 )( 1 mol Pb 2+ /2 mol I – ) = 3.5 x 10-5 f....

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ASA for Expt. 18 1. When PbI 2 (s) is in equilibrium with its saturated solution, [Pb 2+ ][I ] 2 will have a constant value independent of any ions which may be added to, or present in the original solution. 2. a. moles H 2+ = (0.0120 mol/L)(0.00500 L) = 6.00 x 10 -5 b. moles I = (0.0300 mol/L)(0.00500 L) = 1.50 x 10 -4 c. (moles I ) eq = (0.0080 mol/L)(0.010 L) = 8.0 x 10 -5 d. moles I precipitated = 1.50 x 10 -4 – 8.0 x 10 -5 = 7.0 x 10 -5 e. moles Pb 2+ precipitated = (7.0 x 10
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Unformatted text preview: -5 )( 1 mol Pb 2+ /2 mol I – ) = 3.5 x 10-5 f. moles Pb 2+ in sol’n = 6.0 x 10-5 – 3.5 x 10-5 = 2.5 x 10-5 g. [Pb 2+ ] eq = 2.5 x 10-5 mol/0.010 L = 2.5 x 10-3 M h. K sp = [Pb 2+ ][I — ] 2 = (2.5 x 10-3 )( 8.0 x 10-3 ) 2 = 1.6 x 10-7 3. a. [Pb 2+ ] = 1/2[I – ] b. If [I – ] = 5.0 x 10-3 , then [Pb 2+ ] = 1/2(5.0 x 10-3 ) = 2.5 x 10-3 c. K sp = [Pb 2+ ][I — ] 2 = (2.5 x 10-3 )( 5.0 x 10-3 ) 2 = 6.3 x 10-8...
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This lab report was uploaded on 04/08/2008 for the course CHEM 128 taught by Professor All during the Spring '08 term at UConn.

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