Madison Kothari- Equilibrium constant

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Feb 20, 2024

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CHEM1112- section# Experiment #6 Your Name Date Determination of an Equilibrium Constant Lab Partner: Lab Partner’s Name Data and Observations (16 pts) (be mindful of units and significant figures) Part I. Temperature throughout the experiment: Beake r [FeSCN 2+ ] Absorbance at 247 nm 1 1.6*10 -4 Experiment: .615 2 1.2*10 -4 Experiment: .464 3 8*10 -5 Experiment: .340 4 2*10 -5 Experiment: .205 A DVICE FOR FILLING IN THE TABLE . 1) Since in this part of the experiment you used a significant excess of Fe 3+ , amount of product produced ( FeSCN 2+ ) will be limited by the amount of the limiting reactant (SCN - ). That is, [FeSCN 2+ ]=[SCN - ] in this particular case. 2) The moment you mixed your 0.0020 M SCN – solution with other solutions, its concentration (underline) decreased/increased. You can always use M 1 V 1 =M 2 V 2 to calculate molarity changes during dilutions. As long as you use same units for volume on both sides, formula will work. L INEAR R EGRESSION E QUATION 1) Use Microsoft Excel on a lab or personal computer to find an equation that describes data values from your table above. It is commonly known as Beer’s law. Beer’s law y=3385x+0.0675 1

CHEM1112- section# Experiment #6 Your Name Date Determination of an Equilibrium Constant Lab Partner: Lab Partner’s Name Data. Part II Now you’ll be using Beer’s law found above to your benefit, to solve for the unknown [FeSCN 2+ ]. Beaker Absorbance at 457 nm [FeSCN 2+ ] as determined from Beer’s law A Experiment:0.305 7.02*10 -5 B Experiment:0.300 6.87*10 -5 C Experiment:0.272 6.04*10 -5 Show an example of calculation on how you found [FeSCN 2+ ] for the table above: (.305-.0675)/3385=x To find the equilibrium constant (K) value for the formation of [FeSCN 2+ ], you need to know equilibrium concentrations of [Fe 3+ ], [SCN - ], and [FeSCN 2+ ]. You found out [FeSCN 2+ ] from the table above. To find the two remaining equilibrium concentrations, you’ll need to use ICE tables. Advice for filling in the ICE tables below. 1) You can always use M 1 V 1 =M 2 V 2 to calculate initial molarities of Fe 3+ and SCN - . Recall that you diluted their original concentrations, the moment you mixed them with each other and water. As long as you use same units for volume on both sides, formula will work. 2) [Fe(NO 3 ) 3 ] and [Fe 3+ ] are same values, because Fe(NO 3 ) 3 is a soluble salt, a strong electrolyte. In water, it will dissociate completely into Fe 3+ and three times as many NO 3 - , nitrate ions. Therefore, [Fe(NO 3 ) 3 ] = [Fe 3+ ]. Analogously, [KSCN] = [SCN - ]. 3) Equilibrium value of FeSCN 2+ you’ve calculated using Beer’s law, see above. 4) The rest of the table is filled using stoichiometry (logics). Clarify with classmates/instructor to make sure you’re on the right track. 5) After finding the last row of the ICE table (equilibrium concentrations), you can now calculate the equilibrium constant (K) value for this reaction, the key purpose of the lab. BEAKER A Fe 3+ SCN - FeSCN 2+ Initial 6*10 -4 6*10 -4 0.00 Change -7.02*10 -5 -7.02*10 -5 +7.02*10 -5 Equilibrium 5.298*10 -4 5.298*10 -4 7.02*10 -5 2 5 BEAKER B 2

CHEM1112- section# Experiment #6 Your Name Date Determination of an Equilibrium Constant Lab Partner: Lab Partner’s Name Fe 3+ SCN - FeSCN 2+ Initial 6*10 -4 8*10 -4 0.00 Change -6.87*10 -5 -6.87*10 -5 +6.87*10 -5 Equilibrium 5.313*10 -4 7.313*10 -4 6.87*10 -5 K eq = [ FeSCN 2 + ] [ Fe 3 + ][ SCN − ] = 177 BEAKER C Fe 3+ SCN - FeSCN 2+ Initial 6*10 -4 .001 0.00 Change -6.04*10 -5 -6.04*10 -5 +6.04*10 -5 Equilibrium 5.369*10 -4 9.346*10 -4 6.04*10 -5 K eq = [ FeSCN 2 + ] [ Fe 3 + ][ SCN − ] = 120 Did you find K eq to be consistent for all the three reactions? No but they were close Potential Error Sources State at least two inherent sources of error that may have caused inaccuracies in the results and how to minimize or correct them. Be specific. Inherent errors are due to the way the experiment was performed, over which you have little to no control. 1. The curette may not have been completely clean next time we can double clean the curette. 2. There could have also been an air bubble in the solution Postlab Questions (4 pts) 1) Explain in your own words WHY [FeSCN 2+ ]=[SCN ] in the first part of the experiment, but not in the second part of the experiment. The amount of Fe3+ ions is more excessive than the amount of [SCN-] in part 2 the ions are no longer excess it’s the same 2) For this lab, 2 L of 0.0020 M Fe(NO 3 ) 3 needed to be prepared. How many grams of Fe(NO 3 ) 3 needed to be dissolved in water? 0.967g 3

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CHEM1112- section# Experiment #6 Your Name Date Determination of an Equilibrium Constant Lab Partner: Lab Partner’s Name 3) ( bonus question, will bring extra 2 points ). In reality, a hydrate of iron(III) nitrate had to be used, not the anhydrous salt. As you may guess, some of the hydrate’s mass is water, and some is iron(III) nitrate. How many grams of Fe(NO 3 ) 3 • 9H 2 O needed to be dissolved in water to make 2 L of 0.0020 M Fe(NO 3 ) 3 ? Molecular weight of the nonahydrate is 404.0 g/mol. Hint: try to set up an equation using x, and solving it. Assume that the density of your solution is 1.000 g/mL. 1.616g 4