Question

To calculate the equilibrium concentration, you can use the ICE table approach. ICE stands for initial" (1). "change" (C) and equilibrium" (E). The use of ICE tables may have been covered in your lecture. If not, please review the background knowledge posted on iCollege. Answer the following questions. 1. In an experiment of determining the equilibrium constant of ferric thiocyanate, a student group got the absorbance vs [FeSCN*) as shown below. Absorbance vs (FeSCN2+] y=4312x+0.0075 R2=0.9935 Absorbance 0 0000 00001 0.00015 0.0002 0.00025 0.0003 005 Fe(CN) In a test equilibrium system, the student group got an absorbance, 0.32. Initially, they used [Fe'] 0.0010M, [SCN] 0.00060M. Calculate the equilibrium constant of the test system. You can use the following steps to solve the problem. a. Use the absorbance vs [FeSCN] linear equation to determine the concentration of [FeSCN2), when its absorbance is 0.32. b. Use the ICE table to determine the equilibrium concentration of [Fe") and (SCN), [SCN1 [FeSCN2"] [Fe] I initial) C (change) E (equilibrium)

Answer 1

Q1. (a.) From the calibration plot, the equation of straight line is : y = 4312x + 0.0075

where y-axis represents absorbance and x-axis represents equilibrium concentration of [FeSCN²⁺]

Given : absorbance = y = 0.32

From the equation of straight line,

x = (y - 0.0075) / 4312

x = (0.32 - 0.0075) / 4312

x = 7.25 x 10^-5

equilibrium concentration of [FeSCN²⁺] = 7.25 x 10^-5 M

(b.) initial concentration Fe³⁺ = 0.0010 M

initial concentration SCN^- = 0.00060 M

ICE table	[Fe3+]	[SCN-]	$\rightleftharpoons$	[FeSCN2+]
I (initial)	0.0010 M	0.00060 M		0
C (change)	-x	-x		+x
E (equilibrium)	0.0010 M - x	0.00060 M - x		+x

Equilibrium concentration of [FeSCN²⁺] = 7.25 x 10^-5 M (calculated) = x (from table)

Hence, x = 7.25 x 10^-5 M

Equilibrium concentration Fe³⁺ = 0.0010 M - x

Equilibrium concentration Fe³⁺ = 0.0010 M - 7.25 x 10^-5 M

Equilibrium concentration Fe³⁺ = 9.275 x 10^-4 M

Equilibrium concentration SCN^- = 0.00060 M - x

Equilibrium concentration SCN^- = 0.00060 M - 7.25 x 10^-5 M

Equilibrium concentration SCN^- = 5.275 x 10^-4 M

(c.) Equilibrium constant, K = [FeSCN²⁺]_eq / [Fe³⁺]_eq[SCN^-]_eq

K = (7.25 x 10^-5 M) / [(9.275 x 10^-4 M) * (5.275 x 10^-4 M)]

K = 148.2