# ASA16 - -4 = 1.90 x 10 2 2 2 Step 1 Results are as in...

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ASA for Expt. 16 1. Step 1 : mole Fe 3+ = 2.00 x 10 -3 (mol/L) x 0.0050 L = 10.0 x 10 -6< mole SCN - = 2.00 x 10 -3 (mol/L) x 0.0050 L = 10.0 x 10 -6 Step 2 : volume of mixture = 10.0 mL mole FeSCN 2+ = 1.40 x 10 -4 (mol/L) x 0.010 L = 1.40 x 10 -6 Since stoichiometry is 1:1; 1.40 x 10 -6 mol Fe 3+ and SCN - are used up. Step 3 : mole Fe 3+ left = mol SCN - left = 10.0 x 10 -6 - 1.40 x 10 -6 = 8.60 x 10 -6 Step 4 : [Fe 3+ ] = [SCN - ] = 8.60 x 10 -6 /0.010 L = 8.60 x 10 -4 [FeSCN 2+ ] = 1.40 x 10 -6 /0.010 L = 1.40 x 10 -4 volume of solution = 10.0 mL Step 5 : K = [FeSCN 2+ ]/[Fe 3+ ][SCN - ] = 1.40 x 10 -4 /(8.60 x 10 -4 )( 8.60 x 10
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Unformatted text preview: -4 ) = 1.90 x 10 2 2. 2. Step 1 : Results are as in problem 1 Step 2 : mole Fe(SCN) 2 + = 0.70 x 10-4 (mol/L) x 0.010 L = 0.70 x 10-6 mole Fe 3+ = 0.70 x 10-6 mole SCN-= 1.40 x 10-6 Step 3 : mole Fe 3+ = (10.0 x 10-6 - 0.70 x 10-6 ) = 9.3 x 10-6 mole SCN-= (10.0 x 10-6 - 1.40 x 10-6 ) = 8.6 x 10-6 Step 4 : [Fe 3+ ] = 9.3 x 10-6 /0.010 = 9.3 x 10-4 [SCN-] = 8.6 x 10-6 /0.010 = 8.6 x 10-4 [Fe(SCN) 2 + ] = 0.70 x 10-4 Step 5 : K = [Fe(SCN) 2 + ]/[Fe 3+ ][SCN-] 2 = (0.70 x 10-4 )/( 9.3 x 10-4 )( 8.6 x 10-4 ) 2 = 1.0 x 10 4...
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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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