X fesc n 00154 052 0027 2 10 5 m x fesc n 05046 0027

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X FeSC N ¿ + 0.0154 0.52 = 0.027 ¿ 2 + ¿ 10 5 M X FeSC N ¿ 0.5046 = 0.027 ¿ 2 + ¿ 10 5 M 18.6889 = X FeSC N ¿ 4.) 2 + ¿ 10 5 M X FeSC N ¿ + 0.0154 0.56 = 0.027 ¿ 2 + ¿ 10 5 M X FeSC N ¿ 0.5446 = 0.027 ¿ 2 + ¿ 10 5 M 20.1704 = X FeSC N ¿ 5.) 2 + ¿ 10 5 M X FeSC N ¿ + 0.0154 0.78 = 0.027 ¿ 2 + ¿ 10 5 M X FeSC N ¿ 0.7646 = 0.027 ¿ 2 + ¿ 10 5 M 28.3185 = X FeSC N ¿ To find the initial amounts of Fe2+ and SCN- you can use the equation M1V1=M2V2. Solution 1: Fe2+ (.002)(.005)=M2(.01), M2=0.001M SCN- (.002)(.001)=M2(.01), M2=0.0002M Solution 2: Fe2+ (.002)(.005)=M2(.01), M2=0.001M SCN- (.002)(.002)=M2(.01), M2=0.0004M Solution 3: Fe2+ (.002)(.005)=M2(.01), M2=0.001M SCN- (.002)(.003)=M2(.01), M2=0.0006M Solution 4: Fe2+ (.002)(.005)=M2(.01), M2=0.001M SCN- (.002)(.004)=M2(.01), M2=0.0008M
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Solution 5: Fe2+ (.002)(.005)=M2(.01), M2=0.001M SCN- (.002)(.005)=M2(.01), M2=0.001M Now, we can use an ICE table to find Kc for all 5 reactions. Solution 1 Fe3+ SCN- FeSCN2+ Initial Concentration (M) 0.001 0.0002 0 Change in Equilibrium(M) -7.7*10^-5 -7.7*10^-5 7.7*10^-5 Equilibrium Concentration(M) 9.23*10^-4 1.23*10^-4 7.7*10^-5 K c = [ 7.7 10 5 ] [ 9.23 10 4 ][ 1.23 10 4 ] =678.241 Solution 2 Fe3+ SCN- FeSCN2+ Initial Concentration (M) 0.001 0.0004 0 Change in Equilibrium(M) -12.763*10^-5 -12.763*10^-5 12.763*10^-5 Equilibrium Concentration(M) 8.72*10^-4 2.72*10^-4 12.763*10^-5 K c = [ 12.763 10 5 ] [ 8.72 10 4 ][ 2.72 10 4 ] = 538.105 Solution 3 Fe3+ SCN- FeSCN2+ Initial Concentration (M) 0.001 0.0006 0 Change in Equilibrium(M) -18.6889*10^-5 -18.6889*10^-5 18.6889*10^-5 Equilibrium Concentration(M) 8.13*10^-4 4.13*10^-4 18.6889*10^-5 K c = [ 18.6889 10 5 ] [ 8.13 10 4 ][ 4.13 10 4 ] =556.6 Solution 4 Fe3+ SCN- FeSCN2+ Initial Concentration (M) 0.001 0.0008 0 Change in Equilibrium(M) -20.1704*10^-5 -20.1704*10^-5 20.1704*10^-5 Equilibrium Concentration(M) 7.98*10^-4 5.98*10^-4 20.1704*10^-5 K c = [ 20.1704
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  • Fall '06
  • Goddfried
  • Chemistry, Potassium Thiocyanate

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