Experiment Three: The Crystal Violet Experience

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AP Lab Report 3: The Crystal Violet Experience Nathan Duda, Lance Schell, and Laura Wier Park Hill South High School Block One Advanced Placement (AP) Chemistry Tuesday, February 10, 2009 Abstract In this experiment, in order to find the rate of the reaction of crystal violet and OH-, a solution of crystal violet and NaOH was created and placed in a spectrophotometer. The absorbance and percent transmittance were measured at two minute intervals for nineteen minutes. The next step was to graph [Dye] vs. time, ln [Dye] vs. time, and 1 / [dye] vs. time. Since this reaction is first order, the slope of the ln [Dye] vs. time graph is the rate of the reaction. The results of the experiment show that the reaction between crystal violet and NaOH is second order overall, with respect to the dye and OH-. The rate law for the reaction is Rate = k[Dye][OH-]. Introduction The amount of light a substance absorbs is measured in percent transmittance or absorbance. The absorbance values are easier to use as they are directly proportional to the concentration.1 This is the basis of the Beer-Lambert law.1 These values are measured using a spectrophotometer, which allows a certain wavelength of light to pass through a colored medium.2 The colored medium is contained in a cuvette and lowered into the spectrophotometer.2 Distilled water is used first to zero the machine out and establish full transparency values at that particular wavelength.2 In order to equate using the Beer-Lambert law, there are some requirements that must be met, including absorbers which act independently of each other, a homogenous medium (as to prevent diffraction of the radiation), parallel rays must be used in the radiation source, the width of the radiation source

must be more narrow than the medium through which absorbency is being measured, and the light should not cause stimulated emission.3 Depending on the type of experiment being conducted, the rate of formation can be calculated4, various values of an analyte (a substance that is undergoing analysis or is being measured; usually in polarimetry)5, as well as measuring the amount of ultraviolet light absorbed by proteins in a chromatography column6. All measurements are recorded at the substance’s maximum absorbance, which usually varies between wavelengths, and for this reason, measurements of a substance’s absorbance should never be taken only at one wavelength, as extinction exists in the substance at every wavelength in the spectrum.7 However, the approximate middle of the spectrum should be used if only one reading is to be used, as this provides the most accurate measurements.8 Experimental This experiment is called β€œA Study of Reaction Rate” which was made by J. Corsaro.9 The only change made to the experiment was to only measure the percent transmittance and absorbance of the solution for only nineteen minutes instead of the suggested twenty five. Calculations To find the molarity of the dye, use a conversion factor to convert the grams per Liter of dye into moles per liter of dye.

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𝑔 1 π‘šπ‘œπ‘™ 𝐷𝑦𝑒 𝐷𝑦𝑒 Γ— = 6.96 Γ— 10βˆ’5 𝑀 𝐷𝑦𝑒 𝐿 407.99 𝑔 𝐷𝑦𝑒

After the molarity of the dye is found, the concentration of dye can be calculated using the following equation: M1V1 = M2V2

M₁ is the initial molarity of the dye and V₁ is the volume of the dilution. V

is the final

volume of the solution. Use the equation to solve for Mβ‚‚ ([Dye]). See data table 1 (6.96 x 10-5 M)(10 mL) = M2(100 mL) M2 = (6.96 x 10-5 M)(10 mL) (100 mL) M2 = 6.96 x 10-6 To find the concentration of dye in part two of the lab, use the following equation: 𝑦 = π‘šπ‘₯ + 𝑏

The slope and b value is calculated from the graph (see graph 1). Y is the absorbance and x is the unknown [Dye] (See data table 2 for trial one with 10.0mL of stock solution and data table3 for trial two with 5.0mL of stock solution). . 41 = 83945.73 π‘₯ + (βˆ’.046) . 41 βˆ’ (βˆ’.046) =π‘₯ 83945.73 π‘₯ = 5.43 Γ— 10βˆ’6 𝑀

Before the order of the reaction can be determined, the concentration of NaOH for the 5.0mL of stock solution and the 10.0mL of stock solution. First, find the molarity of NaOH using a use a conversion factor to convert the grams per Liter of NaOH into moles per liter of NaOH.

4.2105

𝑔 1 π‘šπ‘œπ‘™ π‘π‘Žπ‘‚π» Γ— = .1053𝑀 π‘π‘Žπ‘‚π» 𝐿 40.00 𝑔 π‘π‘Žπ‘‚π»

Next, use the following equation to determine the molarity of NaOH after dilution:

𝑀₁𝑉₁ = 𝑀₂𝑉₂ . 1053𝑀 10π‘šπΏ = 𝑀(100π‘šπΏ) 𝑀 = .01053𝑀 . 1053𝑀 5π‘šπΏ = 𝑀(100π‘šπΏ) 𝑀 = .005265 𝑀

To find the order of the reaction with respect to NaOH, use the following equation: π‘†π‘™π‘œπ‘π‘’ 𝑇₁ = . 01 π‘₯ π‘†π‘™π‘œπ‘π‘’ 𝑇₂ = . 005 π‘₯

The slopes of trial on and trial two are calculated from the graph (See graph 3 for trial one and graph 5 for trail two). Solve for x. βˆ’30.511 = . 01053 π‘₯ βˆ’16.565 = . 005265 π‘₯ 2 = 2π‘₯ π‘₯=1

X is found to be one, which means that the order of the reaction is first order with respect to NaOH. Conclusion In this experiment, in order to find the rate of the reaction of crystal violet and OH-, a solution of crystal violet and NaOH was created and placed in a spectrophotometer. The absorbance and percent transmittance were measured at two minute intervals for nineteen minutes. The next step was to graph [Dye] vs. time, ln [Dye] vs. time, and 1 / [dye] vs. time. Since this reaction is first order, the slope of the ln [Dye] vs. time graph is the rate of the reaction. The results of the experiment show that the reaction between crystal violet and

NaOH is second order overall, with respect to the dye and OH-. The rate law for the reaction is Rate = k[Dye][OH-]. The following errors may have caused the data to be wrong. In part two, when the 5.0 mL of NaOH was mixed with the 10.0 mL of dye, they values fluctuated and then increased/decreased how they were supposed to. This could be due to the solutions not mixing properly at first, but became mixed and reacted after the first minute.

References 1. Laboratory Activity 1 Teacher Notes. http://intro.chem.okstate.edu/ChemSource/Instrument/inst4.htm 2. Spectrophotometer Use. http://biology.clc.uc.edu/fankhauser/Labs/Microbiology/Growth_Curve/Spectrophotomet er.htm 3. Ebbing, Darrell D, General Chemistry, 1990. 4. IUPAC Gold Book – Rate of Formation vn,y or vc,y. http://goldbook.iupac.org/R05152.html 5. Analyte Definition. Northwestern University Biochemistry Department. http://www.biochem.northwestern.edu/holmgren/Glossary/Definitions/DefA/analyte.html 6. Measurement of Protein Concentration Using Absorbance at 280 nm. http://www.ruf.rice.edu/~bioslabs/methods/protein/abs280.html 7. Using the Spectrophotometer. John Altman. http://www.microbiology.emory.edu/altman/f_protocols/f_instruments/spectrophotomete r.html 8. Setting Up a Colorimetric Assay. David R. Corpette. http://www.ruf.rice.edu/~bioslabs/methods/protein/protcurve.html 9. Corsaro, J., A Study of Reaction Rate, 1964.

Tables Table 1: Crystal Violet Concentration Vs. Absorbance Calibration Curve Dilution (mL) [Dye] %T A 2.0 1.4 x 10-6 M 82% .09 4.0 2.8 x 10-6 M 63% .20 -6 6.0 4.2 x 10 M 52% .28 8.0 5.6 x 10-6 M 42% .37 -6 10.0 6.96 x 10 M 26% .59

Time (min) 1 3 5 7 9 11 13 15 17 19

Table 2: Reaction of Dye With Sodium Hydroxide (NaOH) 5.0 mL NaOH, 10.0 mL Dye %T A [Dye] ln [Dye] 39 .41 5.43 x 10-6 M -12.124 -6 38 .42 5.55 x 10 M -12.102 -6 44 .36 4.84 x 10 M -12.239 -6 47 .33 4.48 x 10 M -12.316 50 .30 4.12 x 10-6 M -12.400 -6 53 .28 3.88 x 10 M -12.460 -6 55 .26 3.65 x 10 M -12.521 56 .25 3.53 x 10-6 M -12.554 -6 57 .24 3.41 x 10 M -12.589 -6 60 .22 3.17 x 10 M -12.662

1 / [Dye] 1.84 x 10-5 M-1 1.80 x 10-5 M-1 2.07 x 10-5 M-1 2.23 x 10-5 M-1 2.43 x 10-5 M-1 2.58 x 10-5 M-1 2.74 x 10-5 M-1 2.83 x 10-5 M-1 2.93 x 10-5 M-1 3.15 x 10-5 M-1

Time (min) 1 3 5 7 9 11 13 15 17 19

Table 3: Reaction of Dye With Sodium Hydroxide (NaOH) 10.0 mL NaOH, 10.0 mL Dye %T A [Dye] ln [Dye] -6 38 .42 5.55 x 10 M -12.102 -6 43 .36 4.84 x 10 M -12.239 49 .31 4.24 x 10-6 M -12.371 -6 55 .26 3.65 x 10 M -12.521 -6 61 .22 3.17 x 10 M -12.662 64 .20 2.93 x 10-6 M -12.741 -6 68 .17 2.57 x 10 M -12.872 -6 71 .15 2.33 x 10 M -12.970 -6 74 .13 2.10 x 10 M -13.074 77 .11 1.86 x 10-6 M -13.195

1 / [Dye] 1.80 x 10-5 M-1 2.07 x 10-5 M-1 2.36 x 10-5 M-1 2.74 x 10-5 M-1 3.15 x 10-5 M-1 3.41 x 10-5 M-1 3.89 x 10-5 M-1 4.29 x 10-5 M-1 4.76 x 10-5 M-1 5.38 x 10-5 M-1

Graphs * The first data point was not plotted to ensure only data which was relevant to the experiment was used. Graph 1: Absorbance vs. [Dye]

Graph 2: [Dye] vs Time

Graph 3: 1 / [Dye] vs. Time

Graph 4: ln [Dye] vs. Time

Graph 5: ln [Dye] vs. Time

(Trial 1)

(Trial 2)

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