Question

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Ne rate of reaction Part 1 Concentration of OH (in bottle) 0-10M Concentration of CV (in bottle) 1.2 x 10M M,V, To find the concentrations of the solutions you created, use M,V, M, is initial concentration of CV V, is the volume of CV you measured out V, is total volume of the final solution Solve for My, the concentration of CV after dilution. Show work. A. Calculate concentration of CV after dilution 1.5 Vol. CV (ml) Vol. of water (ml) Concentration of CV after dilution O bserved Absorbance 1.5 0.100 Beer's Law shows the relationship between absorbance and concentration of a solute in solution Beer's Law: A = abc b A = absorbance of light (collected from instrumentation) E extinction coefficient (how strongly a molecule absorbs light of a particular wavelength) pathlength (the width of the container holding the sample; almost always - 1) C = concentration of the molecule that absorbs light (either known or calculated from experimental data above) The wavelength is set once in this lab and doesn't change, so it's a constant. The pathlength is also constant because the same container is used for every trial. B. Solve for the extinction coefficient. Show work. C. Calculate the concentrations of CV and NaOH (M) for each trial using M,V, = M, V. (You know the reaction occurs almost immediately but calculate the concentrations as though the moment they are mixed, time freezes.) Show work. Vol. of CV (ml) Vol. of water (ml) Vol. of NaOH (ml) Concentration of CV before reaction Concentration of OH before reaction Run 1 0.75 Run 2 0.75 1.5 1.5 0 1. 5H 0.75 1.5 Run 3 0.75

Answer 1

c) i) for concentraron q cu before reactionalkons.. for Runt: TA) Here we have concentrohon o su - we have Mi-1.2 X10M, Y = 0:75 mi V elem 1.2 x16M Yayol.4 CV + JOL Wales volume NACH 60.75 +1.5 +0:15) mi Here you dilute tost by 15ml well and we have to find CODL a bs dilakon mware VEM212 M MIYL 12. M EM212 Volume cut volume weh. 0.3 *105m. M S 1.2*10?m X 075 ml *SAG)3.0ml Sve Emre 0:3 X10SM 1.2x10mx1.5 cm2 x (1:5+1:5)men :: M23.1.5 1.3x18m 30 for Runz. r=1.5.mi ¥23.0m) I M2 I 0.6x105m. _ _ 1: after diterion, we get 0.67165my iv.. . MEMIV - 2*10*4 * 15 ml. 0.6x10me Y2 30mi for Ruri3, 82= zmi. V = 1.5 mi 1.2716 mx .5ml 0.6*105 m 3 ) 70 foro cons..NADH before rachon B) We have, from Beeo's Lamberts 120. AE6C. Puyall given adela AL0109, 63CCM, 10.6 x16MER? :E- Albc 0-100 0.6x105mxiem | 1) por Runtir, = 0.75ml, Vza 3.0ml TEE 16666.67 cmim Ima Miut.0.1 X 0.75 ml. 0.025 m 23.0m1 2) Run 2Y 15. 0 3.0 m M2= o1 mx15ml. 3omo .oSM

3 Runa vic0.75 v2= 3.om mas olimxo-15 mio.025 m 3.0.rar Thys nean reaction. Tcon.cv before come quadhibeft Runi lo.3x105m Jo.025m. kuna 10.6.7105 m l 0.05 m Run 3 10. 6x105 m. 0:025 m 1