Question

Bienvenido a Trip.com In Part 1 of the lab you will make the following four solutions: About 3221 Solution Number 0.500 M Fe3+ 3.00 x 10-4 M SCN' water 2.50 mL 2.50 ml 2.50 mL 2.00 mL 0.50 mL 2.50 mL 1.50 mL 1.00 mL 2.50 mL 1.00 mL 1.50 mL 0 mL Given the data in the table above, determine the initial concentrations of [Fe3+] and [SCN"). Since one of the reactants is in excess, you can determine (FeSCN2+Jeg as well. You will also want to write these in your laboratory notebook as you will need them to determine the value of the molar absorptivity constant for FeSCN2+. Enter the concentrations in the table below. Solution Number [Fe3+Jinitial [SCN Jinitial [FeSCN2+lequil Please enter your results to the correct precision. 3 AM 3 EM 3 Submit Check your answers. Question #2

Answer 1

solution	[Fe3+]initial	[SCN-1]initial	[FeSCN]2-equilibrium
1	0.25M	1.5*10-4 M	1.5*10-4 M
2	0.25M	1.2*10-4 M	1.2*10-4 M
3	0.25M	0.9*10-4 M	0.9*10-4 M
4	0.25M	0.6*10-4 M	0.6*10-4 M

solution 1

before mixing

concentration of Fe³⁺ C₁=0.5 M

volume of Fe^{3+ =} V₁₌2.5 ml

volume of SCN^- ==V_{2 =}2.5 ml

concentration SCN^-1 =C₂ =3*10^-4 M

after mixing

total volume V = volume of Fe³⁺ +volume of SCN^- +volume of water

V= 2.5 ml+2.5ml +0

V =5ml

concentration Fe³⁺ after mixing

using V₁ S₁=V₂S₂ formula

V₁C₁=VC_Fe³⁺

C_Fe³⁺⁼ =V₁C₁ /V

C_Fe³⁺ = 2.5 ml*0.5 M/5ml

C_Fe³⁺ =0.25 M

concentration of SCN^{- after mixing}

V₁C₁=VC_SCN^-

C_SCN^-= =V₁C₁ /V

C_SCN^- = 2.5 ml* 3*10^-4 M/5ml

C_SCN^- =1.5 *10^-4 M

as SCN is present in lesser amount so it will exhaust first so,[FeSCN]²⁺ at equilibrium will be equal to initial concentration of C_SCN^-

[FeSCN]²⁺ at equilibrium =1.2*10^-4 M

for solution 2

before mixing

concentration of Fe³⁺ C₁=0.5 M

volume of Fe^{3+ =} V₁₌2.5 ml

volume of SCN^- ==V_{2 =}2 ml

concentration SCN^-1 =C₂ =3*10^-4 M

after mixing

total volume V = volume of Fe³⁺ +volume of SCN^- +volume of water

V= 2.5 ml+2ml +0.5ml

V =5ml

concentration Fe³⁺ after mixing

using V₁ S₁=V₂S₂ formula

V₁C₁=VC_Fe³⁺

C_Fe³⁺⁼ =V₁C₁ /V

C_Fe³⁺ = 2.5 ml*0.5 M/5ml

C_Fe³⁺ =0.25 M

concentration of SCN^{- after mixing}

V₁C₁=VC_SCN^-

C_SCN^-= =V₁C₁ /V

C_SCN^- = 2 ml* 3*10^-4 M/5ml

C_SCN^- =1.2 *10^-4 M

as SCN is present in lesser amount so it will exhaust first ,[FeSCN]²⁺ at equilibrium will be equal to initial concentration of C_SCN^-

[FeSCN]²⁺ at equilibrium =1.2*10^-4 M

for solution 3

before mixing

concentration of Fe³⁺ C₁=0.5 M

volume of Fe^{3+ =} V₁₌2.5 ml

volume of SCN^- ==V_{2 =1.5} ml

concentration SCN^-1 =C₂ =3*10^-4 M

after mixing

total volume V = volume of Fe³⁺ +volume of SCN^- +volume of water

V= 2.5 ml+1.5ml +1ml

V =5ml

concentration Fe³⁺ after mixing

using V₁ S₁=V₂S₂ formula

V₁C₁=VC_Fe³⁺

C_Fe³⁺⁼ =V₁C₁ /V

C_Fe³⁺ = 2.5 ml*0.5 M/5ml

C_Fe³⁺ =0.25 M

concentration of SCN^{- after mixing}

V₁C₁=VC_SCN^-

C_SCN^-= =V₁C₁ /V

C_SCN^- = 1. 5ml* 3*10^-4 M/5ml

C_SCN^- =0.9 *10^-4 M

as SCN is present in lesser amount so it will exhaust first ,[FeSCN]²⁺ at equilibrium will be equal to initial concentration of C_SCN^-

[FeSCN]²⁺ at equilibrium =0.9*10^-4 M

for solution 4

before mixing

concentration of Fe³⁺ C₁=0.5 M

volume of Fe^{3+ =} V₁₌2.5 ml

volume of SCN^- ==V_{2 =} 1 ml

concentration SCN^-1 =C₂ =3*10^-4 M

after mixing

total volume V = volume of Fe³⁺ +volume of SCN^- +volume of water

V= 2.5 ml+1ml +1.5ml

V =5ml

concentration Fe³⁺ after mixing

using V₁ S₁=V₂S₂ formula

V₁C₁=VC_Fe³⁺

C_Fe³⁺⁼ =V₁C₁ /V

C_Fe³⁺ = 2.5 ml*0.5 M/5ml

C_Fe³⁺ =0.25 M

concentration of SCN^{- after mixing}

V₁C₁=VC_SCN^-

C_SCN^-= =V₁C₁ /V

C_SCN^- = 1 ml* 3*10^-4 M/5ml

C_SCN^- =0.6 *10^-4 M

as SCN is present in lesser amount so it will exhaust first ,[FeSCN]²⁺ at equilibrium will be equal to initial concentration of C_SCN^-

[FeSCN]²⁺ at equilibrium =0.6*10^-4 M