lost on how to do this post lab any help would be appreciated Part I. Analytical

Question

lost on how to do this post lab any help would be appreciated

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Explanation / Answer

From the given experiment

A) Preparation of standard solutions

[SCN-] is the is the limiting reagent here

Total volume of solution = 10 ml

[SCN-] added = [FeSCN]2+ formed

Solution A : [FeSCN]2+ = 9 x 10^-4 M x 3 ml/10 ml = 2.7 x 10^-4 M

Solution B : [FeSCN]2+ = 9 x 10^-4 M x 4 ml/10 ml = 3.6 x 10^-4 M

Solution C : [FeSCN]2+ = 9 x 10^-4 M x 5 ml/10 ml = 4.5 x 10^-4 M

Solution D : [FeSCN]2+ = 9 x 10^-4 M x 6 ml/10 ml = 5.4 x 10^-4 M

Solution E : [FeSCN]2+ = 9 x 10^-4 M x 0 ml/10 ml = 0 M

Plot Absorbance on y-axis versus concentration [FeSCN]2+ on x-axis

Slope = molar extinctional coefficient for [FeSCN]2+

          = (1.83 - 1.088)/(5.4 x 10^-4 - 4.5 x 10^-4)

          = 8244 M-1.cm-1

B) Equilibrium constant

[Fe3+]initial

Solution X : [Fe3+]initial = 0.01 M x 5 ml/10 ml = 0.005 M

Solution Y : [Fe3+]initial = 0.01 M x 3 ml/10 ml = 0.003 M

Solution Z : [Fe3+]initial = 0.01 M x 7 ml/10 ml = 0.007 M

[SCN-]initial

Solution X : [SCN-]initial = 0.0011 M x 5 ml/10 ml = 0.00055 M

Solution Y : [SCN-]initial = 0.0011 M x 7 ml/10 ml = 0.00077 M

Solution Z : [SCN-]initial = 0.0011 M x 3 ml/10 ml = 0.00033 M

[FeSCN]2+eq

Solution X : [FeSCN]2+eq = 0.954/8244 = 1.16 x 10^-4 M

Solution Y : [FeSCN]2+eq = 0.830/8244 = 1.01 x 10^-4 M

Solution Z : [FeSCN]2+eq = 0.600/8244 = 7.28 x 10^-5 M

[Fe3+]eq

Solution X : [Fe3+]eq = [Fe3+]initial - [FeSCN]2+eq = 0.005 - 1.16 x 10^-4 = 4.88 x 10^-3 M

Solution Y : [Fe3+]eq = [Fe3+]initial - [FeSCN]2+eq = 0.003 - 1.01 x 10^-4 = 2.90 x 10^-3 M

Solution Z : [Fe3+]eq = [Fe3+]initial - [FeSCN]2+eq = 0.007 - 7.28 x 10^-5 = 6.93 x 10^-3 M

[SCN-]eq

Solution X : [SCN-]eq = [SCN-]initial - [FeSCN]2+eq = 0.00055 - 1.16 x 10^-4 = 4.34 x 10^-4 M

Solution Y : [SCN-]eq = [SCN-]initial - [FeSCN]2+eq = 0.00077 - 1.01 x 10^-4 = 6.69 x 10^-4 M

Solution Z : [SCN-]eq = [SCN-]initial - [FeSCN]2+eq = 0.00033 - 7.28 x 10^-5 = 2.57 x 10^-4 M

Keq = [FeSCN]2+eq/[Fe3+]eq/[SCN-]eq

Solution X : Keq = (1.16 x 10^-4)/(4.88 x 10^-3)(4.34 x 10^-4) = 54.77

Solution Y : Keq = (1.01 x 10^-4)/(2.90 x 10^-3)(6.69 x 10^-4) = 52.06

Solution Z : Keq = (7.28 x 10^-5)/(6.93 x 10^-3)(2.57 x 10^-4) = 40.90

average Keq = 49.24

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