A STUDENT STUDYING THE FE+3- HSCN equilibrium put into a test tube 10.00ml of 2.00 x 10-3 M Fe(NO3)3 with 10.0 mL of 2.00 x 10-3 M HSCN.THE H+ in the resulting solution was manteined at 0.500.By spectrophotometric analysis of the equilibrium solution, the equilibrium concentration of FESCN+2 was found to be 1.00x10-4M.


1.-CALCULATE THE INITIAL MOLARITY OF THE FE+3 AFTER DILUTION BUT BEFORE REACTION OCCURRED.ie, {Fe+}0
2.- CALCULTE THE INITIAL MOLARITY OF HSCN AFTER DILUTION.
3.- CALCULATE THE FE+3 WHICH REMAINED UNREACTED IN SOLUTION.ie, The equillibrium malarity of Fe +3
4.- CALCULATE THE HSCN MOLARITY WHICH REMAINED UNREACTED IN SOLUTION.ie, The equilibrium molarity.
5.- CALCULATE THE VALUE FOR KC FOR THE REACTION

1. Initial (Fe^3) = 2E-3 x (10 mL/20 mL)= ?M

2. Initial HSCN = 2E-3 x (10 mL/20 mL) = ?

3.
...........Fe^3+ + SCN^- ==> FeSCN^2+
initial...1E-3.....1E-3.......0
change.....-x......-x.........x
equil.......................1E-4
So x = 1E-4. Calculate (F3^^3+) and (SCN^-) and from there plug into Kc expresion and solve for Kc.

To solve this problem, we need to use the balanced chemical equation for the reaction:

Fe3+ + HSCN ⇌ FESCN2+ + H+

The stoichiometry of the reaction is 1:1 between Fe3+ and FESCN2+.

1. To calculate the initial molarity of Fe3+ before the reaction occurred, we can use the dilution formula:
M1V1 = M2V2

Given:
Initial volume of Fe(NO3)3 = 10.00 mL
Initial molarity of Fe(NO3)3 (Fe3+) = 2.00 x 10^-3 M
Final volume of Fe(NO3)3 after dilution = (10.00 mL + 10.0 mL) = 20.00 mL

Let's calculate the initial molarity of Fe3+:
(2.00 x 10^-3 M)(10.00 mL) = (M2)(20.00 mL)
M2 = (2.00 x 10^-3 M)(10.00 mL) / (20.00 mL) = 1.00 x 10^-3 M

So, the initial molarity of Fe3+ before the reaction occurred is 1.00 x 10^-3 M.

2. To calculate the initial molarity of HSCN after dilution, we can use the same dilution formula as before.

Given:
Initial volume of HSCN = 10.0 mL
Initial molarity of HSCN = 2.00 x 10^-3 M
Final volume of HSCN after dilution = (10.0 mL + 10.00 mL) = 20.00 mL

Let's calculate the initial molarity of HSCN:
(2.00 x 10^-3 M)(10.0 mL) = (M2)(20.0 mL)
M2 = (2.00 x 10^-3 M)(10.0 mL) / (20.0 mL) = 1.00 x 10^-3 M

So, the initial molarity of HSCN after dilution is 1.00 x 10^-3 M.

3. To calculate the Fe3+ molarity that remained unreacted in solution (equilibrium molarity of Fe3+), we need to use the initial molarity of Fe3+ and subtract the equilibrium concentration of FESCN2+.

Given:
Equilibrium concentration of FESCN2+ = 1.00 x 10^-4 M

Equilibrium molarity of Fe3+ = Initial molarity of Fe3+ - Equilibrium concentration of FESCN2+
Equilibrium molarity of Fe3+ = (1.00 x 10^-3 M) - (1.00 x 10^-4 M) = 9.00 x 10^-4 M

So, the equilibrium molarity of Fe3+ is 9.00 x 10^-4 M.

4. To calculate the HSCN molarity that remained unreacted in solution (equilibrium molarity), we need to use the initial molarity of HSCN and subtract the equilibrium concentration of FESCN2+.

Given:
Equilibrium concentration of FESCN2+ = 1.00 x 10^-4 M

Equilibrium molarity of HSCN = Initial molarity of HSCN - Equilibrium concentration of FESCN2+
Equilibrium molarity of HSCN = (1.00 x 10^-3 M) - (1.00 x 10^-4 M) = 9.00 x 10^-4 M

So, the equilibrium molarity of HSCN is 9.00 x 10^-4 M.

5. The value of Kc can be calculated using the equilibrium concentrations of FESCN2+, Fe3+, and HSCN, as per the balanced chemical equation.

Given:
Equilibrium concentration of FESCN2+ = 1.00 x 10^-4 M
Equilibrium concentration of Fe3+ = 9.00 x 10^-4 M
Equilibrium concentration of HSCN = 9.00 x 10^-4 M

Kc = [FESCN2+]/([Fe3+][HSCN])
Kc = (1.00 x 10^-4 M) / ((9.00 x 10^-4 M)(9.00 x 10^-4 M))
Kc = 1.23 x 10^4

So, the value of Kc for the reaction is 1.23 x 10^4.

To solve these questions related to the Fe+3-HSCN equilibrium, we need to use the principles of stoichiometry and equilibrium calculations. Let's go through each question step by step:

1. Calculate the initial molarity of Fe+3 after dilution but before the reaction occurred (i.e., [Fe+]0):
To solve this, we need to use the principles of dilution. The initial volume of Fe(NO3)3 is 10.00 mL, and its initial concentration is 2.00 x 10^-3 M.
Using the dilution formula: M1V1 = M2V2, we can calculate the final molarity after dilution:
(2.00 x 10^-3 M) * (10.00 mL) = (X) * (20.00 mL)
Solving for X gives us the initial molarity of Fe+3 (i.e., [Fe+]0).

2. Calculate the initial molarity of HSCN after dilution:
Similar to the previous question, we use the dilution formula: M1V1 = M2V2.
The initial volume of HSCN is 10.0 mL, and its initial concentration is 2.00 x 10^-3 M.
(2.00 x 10^-3 M) * (10.0 mL) = (X) * (20.0 mL)
Solving for X gives us the initial molarity of HSCN (i.e., [HSCN]0).

3. Calculate the Fe+3 which remained unreacted in solution (i.e., the equilibrium molarity of Fe+3):
To determine the equilibrium molarity of Fe+3, we'll subtract the concentration of FESCN+2 at equilibrium (1.00 x 10^-4 M) from the initial molarity of Fe+3 (calculated in question 1).

4. Calculate the HSCN molarity which remained unreacted in solution (i.e., the equilibrium molarity):
Similar to the previous question, the equilibrium molarity of HSCN can be determined by subtracting the concentration of FESCN+2 at equilibrium (1.00 x 10^-4 M) from the initial molarity of HSCN (calculated in question 2).

5. Calculate the value for Kc for the reaction:
The equilibrium constant Kc for the reaction can be calculated using the concentrations of FESCN+2, Fe+3, and HSCN at equilibrium. The equilibrium expression is given by:
Kc = [FESCN+2] / ([Fe+3][HSCN])

Substitute the given values and the equilibrium concentrations obtained in questions 3 and 4 into the equilibrium expression to calculate Kc.

By following these steps, you should be able to calculate the answers to all the provided questions related to the Fe+3-HSCN equilibrium.