Question

1. You prepare a mixture to use to create a colorimetry calibration curve for FeSCN2+ by mixing these things in a 25.00 mL volumetric flask and then diluting to the volume of the flask with 0.3 M HNO3. 5.00 mL 0.00200 M NaSCN 10.00 mL 0.30 M Fe(NO3)3 What is the concentration of FeSCN2+ in this mixture? [Pay attention to the values to determine which reactant is being driven to completely react...]

2. You prepare a mixture to use to measure the equilibrium constant of the reaction of Fe3+ and SCN- ions. You mix these things in a 25.00 mL volumetric flask and then dilute to the volume of the flask with 0.3M HNO3. 3.00 mL 0.00400 M KSCN 10.00 mL 0.00400 M Fe(NO3)3

3.

You prepare a mixture to use to measure the equilibrium constant of the reaction of Fe³⁺ + SCN^- --> FeSCN²⁺

You mix these things in a 25.00 mL volumetric flask and then diluting to the volume of the flask with 0.3 M HNO3
3.00 mL 0.00400 M KSCN
10.00 mL 0.00400 M Fe(NO₃)₃

You use colorimetry to determine that, at equilibrium, the FeSCN²⁺ concentration is 2.4 x 10^-5 M.

Create an I-C-E table and find the K for this particular reaction under these particular conditions

Answer 1

1. Fe(NO₃)₃ is present in large excess; hence, NaSCN is the limiting reactant.

The balanced chemical equation is written as

Fe³⁺ (aq) + SCN^- (aq) ----------> FeSCN²⁺ (aq)

Since SCN^- is the limiting reactant,

[SCN^-] = [FeSCN²⁺]

where the square braces denote molar concentrations.

Use the dilution equation.

M₁V₁ = M₂V₂

where M₁ = molar concentration of stock NaSCN = 0.00200 M; V₁ = volume of stock NaSCN taken = 5.00 mL and V₂ = final volume of the solution = 25.00 mL.

M₂ = concentration of SCN^- in the mixture.

From the dilution equation,

M₂ = M₁V₁/V₂

= (0.00200 M)*(5.00 mL)/(25.00 mL)

= 4.00*10^-4 M.

Concentration of FeSCN²⁺ in the mixture = concentration of SCN^- = 4.00*10^-4 M (ans).

2. Determine the initial concentrations of Fe³⁺ and SCN^- in the test tubes.

[Fe³⁺]_ini = M₁V₁/V₂ = (0.00400 M)*(10.00 mL)/(25.00 mL)

= 1.60*10^-3 M

[SCN^-]_ini = M₁V₁/V₂ = (0.00400 M)*(3.00 mL)/(25.00 mL)

= 4.80*0^-4 M.

Set up the ICE chart for the reaction as below.

Fe³⁺ (aq) + SCN^- (aq) <===========> FeSCN²⁺ (aq)

initial                        1.60*10^-3    4.80*10^-4                                       -

change                       -x                -x                                                +x

equilibrium          (1.60*10^-3 – x)(4.80*10^-4 – x)                              x

It is given that x = 2.40*10^-5 M.

[Fe³⁺]_eq = (1.60*10^-3 – 2.40*10^-5) M = 1.576*10^-3 M.

[SCN^-]_eq = (4.80*10^-4 – 2.40*10^-5) M = 4.56*10^-4 M.

[FeSCN²⁺]_eq = 2.40*10^-5 M.

Write down the expression for the equilibrium constant K as

K = [FeSCN²⁺]_eq/[Fe³⁺]_eq[SCN^-]_eq

= (2.40*10^-5)/(1.576*10^-3)(4.56*10^-4) (ignore units since K is dimensionless)

= 33.39 ≈ 33.4 (ans, correct to 3 sig. figs).