1. You begin preparation of the calibration curve to measure absorbance vs conce

Question

1. You begin preparation of the calibration curve to measure absorbance vs concentration of FeSCN2+. To do so, you add 3.289 mL of 0.200 M Fe(NO3)3 to a cuvette and then directly add 98.989 µL of 0.001 M KSCN. What is the resulting concentration of FeSCN2+, assuming complete conversion of SCN- to FeSCN2+? Provide your response to four digits after the decimal; be careful with your exponent, it must complete this expression: __________ x 10-3M.

2. If the initial amount of Fe(NO3)3 transferred to the cuvette is 0.037 mol, and the absorbance measurements indicate that 0.011 mol of FeSCN2+ are present at equilibrium, what must be [Fe3+] at equilibrium? The total volume of solution in the cuvette is 2.497 mL. Provide your response to two digits after the decimal.

3. Under certain conditions, Kc for the iron thiocyanate system has a value of 134.09. If [Fe3+] = 0.048 M and [SCN-] = 0.04 M, what must be the concentration of FeSCN2+? Provide your response to three digits after the decimal: ____________ M

4. For the iron thiocyanate system, what is the value of the equilibrium constant, Kc, if the following are the concentrations of all species present. Provide your answer to three digits after the decimal.

Explanation / Answer

1) Write the balanced chemical equation for the reaction between Fe3+ and SCN- as below.

Fe3+ (aq) + SCN- (aq) --------> FeSCN2+ (aq)

As per the stoichiometric equation,

1 mole Fe3+ = 1 mole SCN- = 1 mole FeSCN2+

Mole(s) of Fe3+ added = (volume of Fe3+ added in mL)*(concentration on Fe3+ added) = (3.289 mL)*(1 L/1000 mL)*(0.200 M)*(1 mol.L-1/1 M) = 6.578*10-4 mol.

Mole(s) of SCN- added = (volume of SCN- added in L)*(concentration of SCN- added) = (98.989 µL)*(1 L/1.0*106 µL)*(0.001 M)*(1 mol.L-1/1 M) = 9.8989*10-8 mol.

SCN- is the limiting reactant and decides the yield of the product; the amount of FeSCN2+ produced is equal to the amount of SCN- added = 9.8989*10-8 mol.

Total volume of the solution in L = (3.289 mL)*(1 L/1000 mL) + (98.989 µL)*(1 L/1.0*106 µL) = 3.387989*10-3 L.

Concentration of FeSCN2+ = (moles of FeSCN2+)/(total volume of the solution) = (9.8989*10-8 mol)/(3.387989*10-3 L) = 2.92176*10-5 M = (2.92176*10-2)*(10-3 M) = 0.0292176*10-3 M 0.030*10-3 M (ans).

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