# 0.3 Determine the value of an equilibrium constant by complex ion

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Determine the Value of an Equilibrium Constant by Complex Ion Formation by reacting Fe(III) with the thiocyanate ion

Determine the Value of an Equilibrium Constant by Complex Ion Formation

## Objectives

In this laboratory you will:

• Use MicroLab to take colorimetric measurements
• Use Beer’s Law to measure the equilibrium concentration of a complex ion
• Review Le Chatelier’s Principle
• Calculate the equilibrium constant for the formation of a complex ion

• Pre-lab (10%)
• Correctness and thoroughness of your observations and the answers to the questions on the report form (80%)
• TA evaluation of lab procedure (10%)

Before Coming to Lab . . .

• Complete the pre-lab exercise
• Read the introduction and any related materials provided to you

Introduction

When two reactants are mixed, the reaction typically does not go to completion. Rather, they will react to form products until a state is reached whereby the concentrations of the reactants and products remain constant at which point the rate of formation of the products is equal to the rate of formation of the reactants. The reactants and products are then in chemical equilibrium and will remain so until affected by some external force. The equilibrium constant ${K}_{c}$ for the reaction relates the concentration of the reactants and products.

In our experiment we will study the equilibrium properties of the reaction between iron (III) ion and thiocyanate ion:

${\text{Fe}}^{3+}\left(\text{aq}\right)+{\text{SCN}}^{-}\left(\text{aq}\right)\to {\text{FeSCN}}^{2+}\left(\text{aq}\right)$ Equation 1

When solutions containing ${\text{Fe}}^{3+}$ ion and thiocyanate ion are mixed, the deep red thiocyanatoiron (III) ion ( ${\text{FeSCN}}^{2+}$ ) is formed. As a result of the reaction, the starting concentrations of ${\text{Fe}}^{3+}$ and ${\text{SCN}}^{-}$ will decrease: so for every mole of ${\text{FeSCN}}^{2+}$ that is formed, one mole of ${\text{Fe}}^{3+}$ and one mole of ${\text{SCN}}^{-}$ will react. The equilibrium constant expression ${K}_{c}$ , according to the Law of Chemical Equilibrium, for this reaction is formulated as follows:

$\left[{\text{FeSCN}}^{2+}\right]/\left[{\text{Fe}}^{3+}\right]\left[{\text{SCN}}^{-}\right]={K}_{c}$ Equation 2

Remember, square brackets ([]) are used to indicate concentration in mol/liter, i.e. , molarity (M).

The value of ${K}_{c}$ is constant at a given temperature. This means that mixtures containing ${\text{Fe}}^{3+}$ and ${\text{SCN}}^{-}$ will react until the above equation is satisfied, so that the same value of the K c will be obtained no matter what initial amounts of ${\text{Fe}}^{3+}$ and ${\text{SCN}}^{-}$ were used. Our purpose in this experiment will be to find ${K}_{c}$ for this reaction for several mixtures made up in different ways, and to show that ${K}_{c}$ indeed has the same value in each of the mixtures.

The reaction is a particularly good one to study because ${K}_{c}$ is of a convenient magnitude and the red color of the ${\text{FeSCN}}^{2+}$ ion makes for an easy analysis of the equilibrium mixture using a spectrophotometer. The amount of light absorbed by the red complex is measured at 447 nm, the wavelength at which the complex most strongly absorbs. The absorbance, A, of the complex is proportional to its concentration, M, and can be measured directly on the spectrophotometer:

A = kM Equation 3

We know it as the Beer-Lambert law which relates the amount of light being absorbed to the concentration of the substance absorbing the light and the pathlength through which the light passes:

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