1) Frank and Oswalt report a molar absorptivity of 4700 L mol^-1 cm^-1 for thiocyanatoiron(lll) ion. What absorbance would you expect for a soloution that it 1.0e-4 M in thiocyanatorion(lll) ion, if the path length is 1.00 cm?
2) At a given temperature, the equilibrium constant for the system studied in this experiment is 1.40e+2. Suppose 100.0 ml of 2.00e-3 M KSCN was mixed with 100.0 ml of 2.00e-3 M Fe(NO_3_)_3_. At equilibrium, what molar concentration of the thiocyanatoiron(lll) ion would you expect?
A = abc
A = 4700*1*1E-4
Solve for absorbance.
100 mL x 2E-3 = 0.2 millioles Fe^3+
100 mL x 2E-3 = 0.2 millimols SCN^-
(Fe^3+) = 0.2 mmols/200 mL = 0.001M
(SCN^-) = likewise = 0.001M
.......SCN^- + Fe^3+ ==> FeSCN^2+
.....0.001....0.001.......0
C......-x.......-x.........x
E....0.001-x..0.001-x......x
Keq = 140 = (FeSCN^2+)/(SCN^-)(Fe^3+)
Substitute from the E line and solve for x
2.00x10^-3
1) To determine the absorbance of a solution, we can use the Beer-Lambert Law, which states that absorbance (A) is equal to the molar absorptivity (ε) times the concentration (c) times the path length (l):
A = ε * c * l
Given that the molar absorptivity (ε) is 4700 L mol^-1 cm^-1, the concentration (c) is 1.0e-4 M, and the path length (l) is 1.00 cm, we can substitute these values into the equation to find the absorbance:
A = 4700 L mol^-1 cm^-1 * 1.0e-4 M * 1.00 cm
A = 0.47
Therefore, the absorbance of the solution would be 0.47.
2) The equilibrium constant (Kc) for a chemical reaction is a ratio of the concentrations of the products to the concentrations of the reactants. In this case, the thiocyanatoiron(lll) ion is the product, and its molar concentration can be represented as [FeSCN^2+].
Given that the equilibrium constant (Kc) is 1.40e+2, we can set up an expression for the equilibrium concentrations of the thiocyanatoiron(lll) ion:
Kc = [FeSCN^2+]
[KSCN] * [Fe(NO3)3]
Since both KSCN and Fe(NO3)3 have initial concentrations of 2.00e-3 M and the final volume is 200.0 mL (100.0 mL + 100.0 mL), we can set up the equation:
Kc = [FeSCN^2+]
(2.0e-3 M) * (2.0e-3 M)
Simplifying the equation:
1.40e+2 = [FeSCN^2+]
(4.0e-6)
Rearranging the equation to solve for [FeSCN^2+]:
[FeSCN^2+] = (1.40e+2) * (4.0e-6)
[FeSCN^2+] = 5.60e-4
Therefore, at equilibrium, you would expect a molar concentration of the thiocyanatoiron(lll) ion to be 5.60e-4 M.
1) To find the absorbance of the solution, we can use the Beer-Lambert Law, which states that absorbance (A) is directly proportional to the concentration (c) and the path length (l), as well as the molar absorptivity (ε) of the substance:
A = ε * c * l
In this case, they have provided the molar absorptivity (ε) of 4700 L mol^-1 cm^-1, the concentration (c) of 1.0e-4 M, and the path length (l) of 1.00 cm.
Plugging these values into the equation, we get:
A = 4700 L mol^-1 cm^-1 * 1.0e-4 M * 1.00 cm
A = 0.47
Therefore, the absorbance of the solution would be 0.47.
2) The equilibrium constant (K) represents the ratio of the concentrations of products to reactants at equilibrium. In this case, they have provided the equilibrium constant (K) of 1.40e+2.
For the reaction KSCN + Fe(NO3)3 → thiocyanatoiron(lll) ion, we start with 100.0 ml of 2.00e-3 M KSCN and 100.0 ml of 2.00e-3 M Fe(NO3)3.
To determine the molar concentration of the thiocyanatoiron(lll) ion at equilibrium, we need to consider the reaction stoichiometry. From the balanced equation, we can see that 1 mole of KSCN reacts with 1 mole of Fe(NO3)3 to form 1 mole of thiocyanatoiron(lll) ion.
Since we have equal volumes of KSCN and Fe(NO3)3, and their concentrations are the same, the reaction will proceed to produce an equal concentration of the thiocyanatoiron(lll) ion. Therefore, the molar concentration of the thiocyanatoiron(lll) ion at equilibrium would be 2.00e-3 M.
Note: It is important to point out that this conclusion assumes the reaction goes to completion and there are no other side reactions or equilibria present.