help Name Section Experiment 23 Advance Study Assignment: Determination of the E
ID: 877222 • Letter: H
Question
help Name Section Experiment 23 Advance Study Assignment: Determination of the Equilibrium Constant for a Chemical Reaction 1. A student mixes 5.00 ml. 2.00x 10- M Fe(NO,), with 3.00 ml. 2.00 x 10 M KSCN. She finds that in the equiliberium mixture the concentration of FeSCN is 1.28 x 10 M. Find K, for the reaction Fe(aq)+SCN (aq) FeSCN(a) Step 1 Find the number of moles Fel and SCN initially present.(Use Eq 3) moles Fe -moles SCN- How many moles of FeSCN are in the mixture at equilibrium? What is the volume of the equilibrium mixture? (Use Eq. 3.) Step 2 moles FeSCN* How many moles of Fe and SCN- are used up in making the FeSCN"? moles Fe. moles SCN- Step 3 How many moles of Fe and SCN remain in the solution at equilibrium? (Use Eq. 4 and the results of Steps 1 and 2.) -moles Fe", Step 4 What are the concentrations of Fe, SCN-, and FeSCN2+ at equilibrium? What is the volume of the equilibrium mixture? (Use Eq. 3 and the results of Step 3.) mL. Step 5 What is the value of K, for the reaction? (Use Eq 2 and the results of Step 4.) (continwed on following page)Explanation / Answer
Step1 :
From the given data, initial concentration of Fe3+ = (0.005 L) x( 2x10-3 moles/Litre) = 10-5 moles
initial concentration of SCN- = (0.003 L) x( 2x10-3 moles/Litre) = 6 x 10-6 moles
Step 2: At equillibrium one mole of SCN- will react with one mole of Fe3+
Hence No. of moles of FeSCN2+ = 6 x 10-6 moles and Volume = 8 ml
6 x 10-6 moles of Fe3+ and 6 x 10-6 moles of SCN- was used
Step 3: at equillibrium only 4 x 10-6 moles of Fe3+ will remain in the solution
Step 4: At equillibrium Fe3+ concentration = 4 x 10-6 /0.008 M = 5 x 10-4 M, SCN- concentration = 0 and FeSCN2+ concentration = 6 x 10-6 moles/0.008M = 7.5 x10-4 M, Volume = 0.008L
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