Concentration of FeNO33 in 010 M HNO3 solution 0002 M Concen

Concentration of Fe(NO3)3 in 0.10 M HNO3 solution

0.002 M

Concentration of NaSCN in 0.1o M HNO3 solution

0.002 M

Volume of Fe(NO3)3 solution

5.00 mL

Volume of NaSCN solution

2.00 mL

Absorbance

0.346

Equation of the trendline in their graph

y = 2278.6x + 0.0121

1. Calculate the initial [Fe3+] in the test tube.

2. Calculate the initial [SCN^-] in the test tube.

3. Calculate the equilibrium [Fe3+] in the test tube.

4.Calculate the equilibrium [SCN-] in the test tube.

Solution

Use the dilution equation:

M₁*V₁ = M₂*V₂

where M₁ = initial concentration; V₁ = volume of stock solution taken; M₂ = final concentration and V₂ = final volume of the solution.

The total volume of the solution is (5.00 + 2.00) mL = 7.00 mL.

1) Plug in values and obtain

(5.00 mL)*(0.002 M) = [Fe³⁺]_init*(7.00 mL)

====> [Fe³⁺]_init = (5.00*0.002 M)/(7.00) = 0.00143 M (ans).

2) Plug in values and obtain

(2.00 mL)*(0.002 M) = [SCN^-]_init*(7.00 mL)

====> [SCN^-]_init = (2.00*0.002 M)/(7.00) = 0.00057 M (ans).

3) Use the regression equation to find out the concentration of [Fe(SCN)]²⁺ at equilibrium. Put y = 0.346 in the regression equation and obtain

0.346 = 2278.6*[Fe(SCN)²⁺]_eq + 0.0121

=====> 2278.6*[Fe(SCN)²⁺]_eq = 0.346 – 0.0121 = 0.3339

=====> [Fe(SCN)²⁺]_eq = 0.3339/2278.6 = 0.00015 M.

[Fe³⁺]_eq = [Fe³⁺]_init – [Fe(SCN)²⁺]_eq = 0.00143 M – 0.00015 M = 0.00128 M (ans).

4) [SCN^-]_eq = [SCN^-]_init – [Fe(SCN)²⁺]_eq = 0.00057 M – 0.00015 M = 0.00042 M (ans).