Formal lab report .docx - Nadiya Cavallo Date Preformed Lab...

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Nadiya Cavallo Date Preformed: October 11, 2017 Lab Partner: Isaiah Richardson Chemistry 112 Lab Exp. 34 Due Nov. 8, 2017 Experiment 34 Equilibrium Constant Purpose: In this experiment a spectrophotometer was used to test the absorbed radiation in the solutions. The spectrophotometer measured the concentrations of solutions Blank-5 consisting of 0. 2M of Fe(NO 3 ) 3, 0.001 M of NaSCN, and 0.1 M of HNO 3 , and solutions 6-10 consisting of 0.002 M of Fe(NO 3 ) 3 , 0.002 M NaSCN, and 0.1 M of HNO 3 and analyzed the absorption rate to find the constant K. Theory: In theory we will be determining the equilibrium constant K c . Equilibrium is when the reactants and the products are equal in concentrations and the forward and reverse reaction are happening at an equal rate. K c is the equilibrium constant at a given temperature that we will be using in this experiment and it tells us where the equilibrium point is reached. We will be using a spectrophotometer to collect data to calculate K, spectrophotometry, according to the lab book, is measuring light intensities with a photosensitive detector at specific visible wavelengths. Certain aspects will affect the absorbance like the concentration, thickness of the sample, and the
light absorption. We will be focusing on the thickness and molar concentration of the absorbing substance. Procedure: At the beginning of lab we turn in all out work form the previous class and I we are not already in proper clothing then we get changed. We grabbed goggles, 11 large test tubes, 5 50 ml beakers, 2 pipet bulbs, and 1,2,3,4, 5, and 10 ml pipets. We then labeled the test tubes blank – 10.
In part A (Blank – 5) we prepared 5 solutions and a blank in 25 mL volumetric flasks. The blank consisted of 10.0 mL of 0.2 M Fe(NO 3 ) 3 , 0 mL of 0.001 M NaSCN, and the diluted it to 25 ml with HNO 3 . Then we added 1, 2, 3, 4, and 5 ml of 0.001 M of NaSCN to the remaining test tubes. Test tubes 1-5 then all got 10.0 ml of 0.2 M of Fe(NO 3 ) 3 and then were each diluted with 0.1 M of HNO 3 so that they all equaled 25 ml.

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