5. In this experiment, you will prepare a standard solution for a calibration cu

Question

5. In this experiment, you will prepare a standard solution for a calibration cur formation of FeSCN+ complex by reaction different volumes of 0.002M Fe(NO. 0.002 M NasCN. You will use 0.1M HNO, for all dilutions. Using the table, Calculate the equilibrium concentrations of FeSCN2+. Vol of Fe3+ (mL) Vol of SCN- (mL) 0.1 M HNO3 (mL) Equilibrium concentration Fe[SCN]2+ 10.00 0.00 10.00 10.00 1.00 9.00 10.00 2.00 8.00 10.00 3.00 7.00 10.00 4.00 6.00 10.00 5.00 5.00 Tent 12 | Le Chatelier's Principle: Chemical Equilibrium and Determination of Equilibrium Constant: Pre-lab 181 -mi-

Explanation / Answer

In this experiment we will study the equilibrium properties of the reaction between the iron(III) ion and thiocyanic acid (HSCN):

Fe3+ (aq) + HSCN (aq) ¥ FeSCN2+ (aq) + H+ (aq)

When solutions containing Fe3+ ion and thiocyanic acid are mixed, reaction occurs to some extent, forming the FeSCN2+ complex ion, which has a deep blood red color, and H+ ion. As a result of the reaction, the equilibrium amounts of Fe3+ and HSCN will be less than they would have been if no reaction had occurred; for every mole of FeSCN2+ that is formed, one mole of Fe3+ and one mole of HSCN will react.

According to the general law, Keq for reaction takes the following form.

Keq = [Fe ][HSCN] / [FeSCN ][H ]

Our purpose in the experiment will be to evaluate Keq for the reaction by determining the equilibrium concentrations of the four species in equation in several solutions made up in different ways. The equilibrium constant for the reaction has a convenient magnitude and the color of the FeSCN2+ ion makes for an easy analysis of the equilibrium mixture. The solutions will be prepared by mixing solutions containing known concentrations of iron(III) nitrate and thiocyanic acid. The color of the FeSCN2+ ion formed will allow us to determine its equilibrium concentration by spectroscopy. Knowing the initial composition of the solution and the equilibrium concentration of FeSCN2+, we can calculate the equilibrium concentrations of the rest of the pertinent species and then calculate Keq.

Since the calculations that are necessary to find =Keq may not be apparent, let us consider a specific example. Assume that we prepare our solution by mixing 10.0 mL of 2.00 x 10-3 M Fe(NO3)3 with 10.0 mL of 2.00 x 10-3 M HSCN under conditions which keep [H+ ] equal to 0.100 M.

The Fe3+ in the iron(III) nitrate reacts with the HSCN to produce some red FeSCN2+ complex ion. By spectroscopy and Beer’s Law, it is found that [FeSCN2+] at equilibrium is 1.50 x 10-4 M.

To find Keq for the reaction from these data, it is convenient first to determine how many moles of reactant species were initially present before a reaction occurred. By the definition of the molarity of a species A,

moles A = MAV where V is the volume of the solution in liters. (Remember that 1000 mL = 1 L.)

initial moles Fe3+ = (MFe 3+)(VFe 3+) = 2.00 x 10-3 mol/L (0.0100 L) = 2.00 x 10-5 mol Fe 3+

initial moles HSCN = (MHSCN)(VHSCN) = 2.00 x 10-3 mol/L (0.0100 L) = 2.00 x 10-5 mol HSCN

The number of moles of FeSCN2+ present at equilibrium is found from the molarity and the volume of the solution (10.0 mL + 10.0 mL = 20.0 mL).

equilibrium moles FeSCN2+ = (MFeSCN 2+)(VFeSCN 2+) = 1.50 x 10-4 M (0.0200 L) = 3.00 x 10-6 mol FeSCN2+ The FeSCN2+ ion is produced as shown in equation

Therefore, for every mole of FeSCN2+ present in the equilibrium mixture, one mole Fe3+ and one mole HSCN are reacted.

We can see then that equilibrium moles Fe3+ = initial moles Fe3+ – equilibrium moles FeSCN2+

equilibrium moles Fe3+ = 2.00 x 10-5 mol – 3.00 x 10-6 mol = 1.70 x 10-5 mol Fe 3+

Similarly for HSCN, equilibrium moles HSCN = 2.00 x 10-5 mol – 3.00 x 10-6 mol = 1.70 x 10-5 mol HSCN Knowing the number of moles of Fe3+ and HSCN present in the equilibrium mixture and the volume of the mixture, we can easily find the concentrations of those two species.

Fe+3= 1.70 x 10 mol/0.002 L = 8.50 x 10-4 M

HSCN= 1.70 x 10 mol/0.002 L = 8.50 x 10-4 M

Keq =[ FeSCN ] [ H] / [Fe] [HSCN] = 1.50 * 10-4*0.100 /  8.50 x 10-4 M* 8.50 x 10-4 M = 20.6

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