Question

Chemistry (3 part Questions)

3a. Use the following data table and equation to generate a standard curve for the concentration of the product [Fe(SCN)]⁺² by plotting equilibrium concentration versus absorbance. A graph is required for credit including title, labeled axis, and correct linear regression line(s) ?

Fe⁺³ + SCN^-1 <--> [Fe(SCN)]⁺²

Test Tube	0.100 M FeCl3	0.00100M KSCN	Water	Absorbance
1	10.00 mL	1.00 mL	9.00 mL	0.012
2	10.00	2.00	8.00	0.101
3	10.00	3.00	7.00	0.245
4	10.00	4.00	6.00	0.338
5	10.00	5.00	5.00	0.456
6	10.00	6.00	4.00	0.557
7	10.00	7.00	3.00	0.793
8	10.00	8.00	2.00	0.848

3b.Calculate the equilibrium concentration of a sample of [Fe(SCN)]⁺² with an absorbance of 0.194 ?

3c. Calculate the equilibrium constant for a reaction with an initial concentration of 0.178 M Fe⁺³ and 0.153 M SCN^-1 with an absorbance at the equilibrium of 0.194 ?

I have no idea how to do this problem, the first part needs a graph (Step by step on how u label and graph it) and the 2nd and 3rd part Idk how to do it. Please show a graph and do a step by step on how to solve for the answer. Your help is always greatly appreciated! Thank you in advance.

Answer 1

For the equillibrium reaction:

Fe³⁺ + SCN^- [Fe(SCN)]²⁺

let us assume that whole of the limiting reagent is consumed during the reaction. Since product is a colored compound, will show absorbance in visivle range. From the given concentrations it is apparent that SCN^- solution has much lower concentration in comparison to Fe³⁺. therefore it is assumed that whole of SCN^- will react and moreover the stoichiometric ratio of SCN^- to product is 1:1, hence same concentration of product will form .i.e initial [SCN^-] = equillibrium concentration of [Fe(SCN)]²⁺

Step 1: Calculate the concentration of reactants in the reaction mixture as:

M₁*V₁ = M₂*V₂

where M₁ nad M₂ are the concentration of stock solution used and concentration in reaction mixture, respectively

V₁ and V₂ are the volumes of stock solution used and final (total) volume of the reaction mixture = 20 mL (from the table provided)

example or test tube 1:

0.1*10=M₂*20

or [Fe³⁺] = 0.05M

Similarly fot [SCN^-]:

0.001*1=M₂*20

or [SCN^-] = 5*10^-5 M (therefore its the limiting reagent in the reaction)

See the table below for all the test solutions and corresponding plot.

3 b. Calculate the equilibrium concentration of a sample of [Fe(SCN)]⁺² with an absorbance of 0.194 ?

Using the Lambert-Beer's Law as:

A = Cl

where A=absorbance, = absorptivity coefficient, C = concentartion of [Fe(SCN)]⁺² , l=path length (=1 cm)

From the above plot of Absorbance vs concentration and fitting it to a linear function as y=mx+c where m = slope = el = 20000 and

c = intercept = 0. Therefore is calculated as:

* 1 cm=20000 M^-1

= 20000 M^-1cm^-1

Now using the same Lambert-Beer's Law and A= 0.194, = 20000 M^-1cm^-1 and l= 1 cm, C is calculated as:

0.194 = 20000 *C*1

or C= 9.7 *10^-6 M

Therefore, the equilibrium concentration of a sample of [Fe(SCN)]⁺² with an absorbance of 0.194 is 9.7 * 10^-6 M

3c. Calculate equilibrium constant for a reaction with an initial concentration of 0.178 M Fe⁺³ and 0.153 M SCN^-1 with an absorbance at the equilibrium of 0.194.

Ans: For the reaction equillibrium constant , k_eq = [Fe(SCN)]⁺²/[Fe⁺³]*[SCN^-1]

where concentrations of reactants and product are at equillibrium

from part b the equilibrium concentration of a sample of [Fe(SCN)]⁺² with an absorbance of 0.194 is 9.7 * 10^-6 M.

Here the concentration of reactants are comparable thus, there equillibrium concentartions are calculated as follows:

From absorbance values, [Fe(SCN)]⁺² = 9.7 * 10^-6 M =x

therefore, at equillibrium [Fe⁺³] = 0.178-x = 0.178 - 9.7 * 10^-6 M

and [SCN^-1] = 0.153-x = 0.153 - 9.7 * 10^-6 M

substituting in above formula of equillibrium constant we get:

k_eq = 9.7 * 10^-6/[(0.178 - 9.7 * 10^-6)*(0.178 - 9.7 * 10^-6)]

or k_eq= 3.56 * 10^-4

Therefore, equilibrium constant for a reaction with an initial concentration of 0.178 M Fe⁺³ and 0.153 M SCN^-1 with an absorbance at the equilibrium of 0.194 is 3.56 * 10^-4 .