Question

3. In Part A, you measure the absorbance of four FesCN" solutions and prepare a plot (Absorbance vs. concentration) similar to the one below. Beer's Law Plot for FeSCN2 0.6 0.5 0.4 0.3 d 0.2 0.1 0 y = 2417.5× + 0.0483 R2 0.99897 0.00025 0.00015 0.0002 0.00005 0.0001 0 Concentration (M) You use this plot as a standard curve in Part B to determine the concentration of an unknown. If an unknown FeSCN2 solution gives an absorbance reading of 0.334, use the equation of the line shown above to determine its concentration. Show all your work below.

Answer 1

if

A = 0.334

then

A = 2417.5*C + 0.0483

0.334= 2417.5*C + 0.0483

solve for C

C = (0.334-0.0483)/2417.5

C = 0.000118 M

[FeSCN+2] = 0.000118 M

Q4

Kf = [FeSCN+2]/[Fe+3][SCN-]

initially

[Fe+3] = M1V1/(Vtotal) = 0.002*3/(3+2+5) = 0.0006

[SCN-] = M1V1/(Vtotal) = 0.002*2/(3+2+5) = 0.0004

[FeSCN+2] = 0

in equilbirium

[Fe+3] = = 0.0006-x

[SCN-] = = 0.0004 - x

[FeSCN+2] = x

get x

A = 2417.5*C + 0.0483

0.118 = 2417.5*C + 0.0483

C =( 0.118 -0.0483)/2417.5 = 0.0000288 M

[Fe+3] = = 0.0006-0.0000288 = 0.0005712

[SCN-] = = 0.0004 - 0.0000288 = 0.0003712

[FeSCN+2] = 0.0000288

substitute in Kf

Kf = (0.0000288 )/(0.0003712*0.0005712 )

Kf = 135.83