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.
2.- CALCULTE THE INITIAL MOLARITY OF HSCN AFTER DILUTION.
3.- CALCULATE THE FE+3 WHICH REMAINED UNREACTED IN SOLUTION.
4.- CALCULATE THE HSCN MOLARITY WHICH REMAINED UNREACTED IN SOLUTION.
5.- CALCULATE THE VALUE FOR KC FOR THE REACTION.
I HOPE YOU CAN DO THIS HELP PLEASE IM LOST
1. To calculate the initial molarity of Fe+3 after dilution but before the reaction occurred, we can use the equation for dilution:
M1V1 = M2V2
Where:
M1 = initial molarity of Fe(NO3)3 = 2.00 x 10^-3 M
V1 = initial volume of Fe(NO3)3 = 10.00 mL = 0.0100 L (converted to liters)
M2 = final molarity of Fe+3 after dilution = to be determined
V2 = final volume of Fe+3 after dilution = 20.00 mL + 10.00 mL = 30.00 mL = 0.0300 L (converted to liters)
Substituting the values into the equation, we can solve for M2:
(2.00 x 10^-3 M)(0.0100 L) = M2(0.0300 L)
M2 = (2.00 x 10^-3 M)(0.0100 L) / 0.0300 L
M2 ≈ 6.67 x 10^-4 M
Therefore, the initial molarity of Fe+3 after dilution but before the reaction occurred is approximately 6.67 x 10^-4 M.
2. The initial molarity of HSCN after dilution is the same as the initial molarity of Fe(NO3)3, which is 2.00 x 10^-3 M.
3. To calculate the Fe+3 which remained unreacted in solution, we need to subtract the equilibrium concentration of FESCN+2 (1.00 x 10^-4 M) from the initial molarity of Fe+3 after dilution (6.67 x 10^-4 M):
Fe+3 remaining = 6.67 x 10^-4 M - 1.00 x 10^-4 M = 5.67 x 10^-4 M
Therefore, the Fe+3 which remained unreacted in solution is 5.67 x 10^-4 M.
4. Similarly, the HSCN molarity which remained unreacted in solution is equal to the initial molarity of HSCN (2.00 x 10^-3 M) minus the equilibrium concentration of FESCN+2 (1.00 x 10^-4 M):
HSCN remaining = 2.00 x 10^-3 M - 1.00 x 10^-4 M = 1.90 x 10^-3 M
Therefore, the HSCN molarity which remained unreacted in solution is 1.90 x 10^-3 M.
5. The equilibrium constant (Kc) for the reaction can be calculated using the equation:
Kc = [FESCN+2] / ([Fe+3][HSCN])
Substituting the given values into the equation:
Kc = (1.00 x 10^-4 M) / ((5.67 x 10^-4 M)(1.90 x 10^-3 M))
Simplifying the expression:
Kc = 1.00 x 10^-4 M / (1.08 x 10^-6 M^2)
Kc ≈ 92.6 M^-1
Therefore, the value for Kc for the reaction is approximately 92.6 M^-1.
Sure, I can help you with that! Let's break down each question step by step:
1. To calculate the initial molarity of Fe+3 after dilution but before the reaction occurred, we need to consider the volumes and concentrations of the solutions. The total volume of the solution after dilution is 10.00 mL (Fe+3 solution) + 10.0 mL (HSCN solution) = 20.0 mL = 0.020 L. The initial molarity of Fe+3 before dilution is 2.00 x 10^-3 M.
To find the initial molarity after dilution, we can use the dilution equation:
M1V1 = M2V2
M1 = initial molarity
V1 = initial volume
M2 = final molarity
V2 = final volume
In this case, the initial molarity (M1) of Fe+3 is 2.00 x 10^-3 M, and the initial volume (V1) is 10.00 mL = 0.010 L. The final volume (V2) is 0.020 L, and we can solve for the final molarity (M2) of Fe+3:
(2.00 x 10^-3 M)(0.010 L) = M2(0.020 L)
M2 = (2.00 x 10^-3 M)(0.010 L) / (0.020 L)
= 0.001 M
Therefore, the initial molarity of Fe+3 after dilution but before the reaction occurred is 0.001 M.
2. Similar to the previous question, we can calculate the initial molarity of HSCN after dilution using the dilution equation. The initial molarity (M1) of HSCN is also 2.00 x 10^-3 M, and the initial volume (V1) is 10.0 mL = 0.010 L. The final volume (V2) is 0.020 L. Solving for the final molarity (M2) of HSCN:
(2.00 x 10^-3 M)(0.010 L) = M2(0.020 L)
M2 = (2.00 x 10^-3 M)(0.010 L) / (0.020 L)
= 0.001 M
Therefore, the initial molarity of HSCN after dilution is also 0.001 M.
3. To calculate the Fe+3 which remained unreacted in solution, we can use the balanced equation for the reaction:
Fe+3 + HSCN ⇌ FESCN+2 + H+
The stoichiometry of the reaction is 1:1, which means that for every 1 mole of Fe+3, there will be 1 mole of FESCN+2 formed. Since the equilibrium concentration of FESCN+2 is given as 1.00 x 10^-4 M, the amount of Fe+3 reacted is also 1.00 x 10^-4 M.
The initial molarity of Fe+3 after dilution is 0.001 M (as calculated in question 1). Therefore, the Fe+3 which remained unreacted in solution is:
0.001 M - 1.00 x 10^-4 M
= 9.00 x 10^-4 M
4. Similarly, to calculate the HSCN molarity which remained unreacted in the solution, we need to consider the stoichiometry of the reaction. Since the stoichiometry is 1:1, the initial molarity of HSCN after dilution (0.001 M) will be equal to the HSCN molarity which remained unreacted.
Therefore, the HSCN molarity which remained unreacted in solution is 0.001 M.
5. The value for Kc can be calculated using the equation:
Kc = ([FESCN+2]/[Fe+3])([H+]/[HSCN])
From the information given in the question, we know that the equilibrium concentration of FESCN+2 is 1.00 x 10^-4 M, the concentration of H+ is 0.500 M (maintained at 0.500), and the concentration of HSCN (which remained unreacted) is 0.001 M.
Plugging in these values into the equation, we can calculate Kc:
Kc = (1.00 x 10^-4 M / 9.00 x 10^-4 M)(0.500 M / 0.001 M)
= 1.11 x 10^-1
Therefore, the value for Kc for the reaction is 1.11 x 10^-1.
I hope this explanation helps, and let me know if you have any further questions!