After mixing 5.00 mL of the 5.0×10–5 M crystal violet solution with 5.00 mL of 0.20...

After mixing 5.00 mL of the 5.0×10–5 M crystal violet solution with 5.00 mL of 0.20 M NaOH, what is the concentration of crystal violet and hydroxide ions in the solution at the start of the reaction? Is it a reasonable assumption that the concentration of OH– is much greater than that of CV+? Why is this an important assumption? Explain.

Expert Solution

queen_honey_blossom answered 1 year ago

1. A student mixes 3.00 mL of 2.27 × 10−5 M crystal violet solution with 3.00...

1. A student mixes 3.00 mL of 2.27 × 10−5 M crystal violet solution with 3.00 mL 0.100 M sodium hydroxide both at 24.5 °C and collects the following data: Time(Min) Absorbance 1.49 0.206 4.17 0.157 6.22 0.131 8.15 0.108 10.10 0.093 13.00 0.072 14.02 0.066 16.18 0.055 19.00 0.044 (a) Describe the graphical analysis steps the studentshould perform in order to determine the (i) Order of the reaction with respect to crystal violet (ii) Value of the rate constant,...

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Calculate the molar absorptivity of crystal violet using the absorbance value of the crystal violet solution...

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Concentration of NaOH solution= 0.02803 M Concentration of Crystal Violet dye solution=0.001 M Room Temperature =25...

Concentration of NaOH solution= 0.02803 M Concentration of Crystal Violet dye solution=0.001 M Room Temperature =25 degree celcius Absorbance from the device- spectrovis(spectrophotometer) The data in the trials are the absorbance. -Its not Organic chemistry. Its an experiment about the crystal violet dye. Mainly they want to know how we get the rate law using these numbers. It's asking for the rate law through these 2 trails and its asking to use some numbers to show how its solved . I...

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Information regarding the question: Concentration of NaOH=0.05138 M Concentration of Crystal Violet Dye Solution=0.001 M Room...

Information regarding the question: Concentration of NaOH=0.05138 M Concentration of Crystal Violet Dye Solution=0.001 M Room Temperature=25 degree Celcius Absorbance from the device- spectrovis(spectrophotometer) Time: Trial 1 Trial 2 0 0.089 0.430 30 0.106 0.323 60 0.121 0.248 90 0.106 0.187 120 0.063 0.141 150 0.045 0.109 180 0.028 0.082 210 0.023 0.060 240 0.014 0.045 270 0.010 0.036 300 0.005 0.027 330 0.003 0.017 360 0.000 0.015 390 -0.001 0.008 420 -0.002 0.006 450 -0.002 0.007 The...

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A solution is prepared by mixing 150.0 mL of 0.200 M Mg2+ and 250.0 mL of 0.100 M F-. A precipitate forms. Calculate the concentrations of Mg2+ and F- at equilibrium. Ksp = 6.4 × 10-9

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imagine mixing 25.0 ml of 0.10 M acetic acid (Ka=1.8*10^-5) with 25.0 ml of 0.10 M sodium acetate and determine a) The initial pH of the buffer b) pH of 20.0 ml sample of buffer with 1.0 ml of 0.10M Hal added c) pH of another 20 mL sample of buffer with 1.0 ml of 0.10 M MaOH added

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