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question 3 please

Integrated Chemical Kinetics Determination of the Rate Law for the Crystal Violet/Hydroxide ion reaction Page 7 Pre-laboratory Assignment Name: Show all work in your lab notebook and box your final answers, Express to correct sig figs and include units 1) 7.00 mL of 0.00750 M CV solution was diluted to 10,0 mL. Calculate the Beer's law constant Ks if the diluted solution's absorbance was measured to be 0.275 using a spectrophotometer. 2) 25.0 mL of a solution prepared as 0.00750 M CV is added to 55.0 mL of solution prepared as 0.115 M NaOH. Calculate the initial concentrations of both CV and OH in the reaction mixture. 3) The experiment stated in problem 2 was performed and absorbance of CV' was measured over time. Calculate the concentration, M of CV' in the reaction mixture for each time interval shown. (Note: You must use the information stated in problem 1). Time(min) Conc., M Absorbance 0.123 0.118 0.100 1.00 2.00

Answer 1

Integrated Chemical Kinetic Determination of the Rate Law for the Crystal Violet/Hydroxide ion reaction Pre-laboratory Assignment Name: Show all work in your la notebook and box your final answers. Express to correct tigfigs and include units. 1 7.00 ml of 0.00750 M CV solution was diluted to 100 ml. Calculate the Beer's law constant kif the diluted solution's absorbance was measured to be 0.275 using a spectrophotometer (MiN,) = (M_V3_) ( :-Dilution) 0.00750x 7 = 4 x 10 0.00525M=M2 A=k, C (Lambert - Lawre) 0.275 = kb X0.00 525 2) 250 ml of a solution prepared as 0.00750 M CV is added to 55.0 mL of solution prepared as 0.115 M NaOH. Calculate the initial concentrations of both CV and OH in the reaction mixture 3) The experiment stated in problem 2 was performed and absorbance of CV was measured over time. Calculate the concentration, M of CV in the reaction mixture for each time intervalshown. (Note: You must use the information stated in problem 1). Time min) Conc. M Absorbance 0.123 0.118 0.100 1.00 2.00 K6= 0.275 0.00525 1 K6=52.38

At Time o min O At time 2 min A = Koc 0.123 = 52.38 XC C= 0.002348 MI 0.100 = 52.38 XC Ic=0.00 1909 M/ At Time (min - ке с 0.118 = 52.38 X C TC= 0.00 2252 MS