Astandard solution of fescn2 is prepared by combining 9.0 ml of 0.20 m fe(no3)3 with 1.0 ml of 0.0020 m kscn. the standard solution had an absorbance of 0.510. a trial solution was made in a similar manner, but with a more dilute fe(no3)3 reagent. the initial scn– concentration (immediately after mixing) was 0.00080 m. this trial solution had absorbance of 0.240. what is the equilibrium concentration of scn– in the trial solution?

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