Respuesta :
Answer:
The equilibrium concentration of SCN- in the trial solution is 0.0007059 M
Explanation:
Step 1: The equation
Fe(NO3)3 + KSCN ==> FeSCN^2+ + 3NO3^- +K+
Step 2: First we have to calculate the initial moles of Fe3+ and SCN-
Moles of Fe3+ = Concentration of Fe3+ * Volume of the solution
Moles of Fe3+ = 0.2M * 9*10-3 L = 0.0018 moles
Moles of SCN- = Concentration of SCN- * volume of the solution
Moles of SCN- = 0.0020 M* 1*10^-3 = 0.000002 moles
When 1 mole of Fe3+ is consumed, there is needed 1 mole of SCN- to produce 1 mole of FeSCN2+
SCN- os the limiting reagens, this means it will completely react.
Since there is 1 mole of SCN- needed to produce 1 mole of FeSCN2+
So if there is consumed 0.000002 moles of SCN-, there is also produced 0.000002 mole of FeSCN2+
Step 3: Calculate the concentration of FeSCN2+
Concentration of FeSCN2+ = 0.000002 mole of FeSCN2+/ (9+1 *10^-3L) = 0.0002 M
Step 4: using Lambert-beer's law:
A = c*C*l
with A= the absorbance of the solution
with C = concentration of the solution
with l= the path length
with c = molar absorptivity coefficient
c and l are same for stock solution and dilute solution.
c*l = A/C = 0.510/0.0002M = 2550M^-1
For the trial solution:
The equilibrium concentration of SCN^- is:
[SCN-]eq = [SCN-]initial - [FeSCN2+]
Step 5: Calculate concentration of FeSCN2+
C= A/cl
C = 0.240/2550 = 9.41*10 ^-5 M
Step 6: Calculate concentration of SCN-
[SCN-]eq = [SCN-]initial - [FeSCN2+]
[SCN-]eq = 0.00080 M - 9.41*10 ^-5 M
[SCN-]eq = 0.0007059 M
The equilibrium concentration of SCN- in the trial solution is 0.0007059 M
