a. Percent transmittance for an indicator was measured at 590 nm at various pHs. Determine the pK. for the indicator. pH 2.35 3.45 4.42 5.40 6.60 7,88 11.05 %T 94.31 44.50 33.40 25.50 18.45 13.40 7.89 b. What color is the indicator at pH 11.05?
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- The acid-base indicator HIn undergoes the following reaction in dilute aqueous solution: HIncolor1H++Incolor2 The following absorbance data were obtained for a 5.00 I0-4 M solution of HIn in 0.1 M NaOH and 0.1 M HC1. Measurements were made at wavelengths of 485 nm and 625 nm with 1.00-cm cells. 0.1 M NaOH A485 = 0.075 A625 = 0.904 0.1 M HC1 A485 = 0.487 A625 = 0.181 In the NaOH solution, essentially all of the indicator is present as In-; in the acidic solution, it is essentially all in the form of HIn. (a) Calculate molar absorptivities for In- and HIn at 485 and 625 nm. (b) Calculate the acid dissociation constant for the indicator ¡fa pH 5.00 buffer containing a small amount of the indicator exhibits an absorbance of 0.567 at 485 nm and 0.395 at 625 nm (1.00-cm cells). (c) What is the pH of a solution containing a small amount of the indicator that exhibits an absorbance of0.492 at 485 nm and 0.245 at 635 nm (1.00-cm cells)? (d) A 25.00-mL aliquot of a solution of purified weak organic acid HX required exactly 24.20 mL of a standard solution of a strong base to reach a phenolphthalein end point. When exactly 12.10 mL of the base was added to a second 25.00-mL aliquot of the acid, which contained a small amount of the Indicator under consideration, the absorbance was found to be 0.333 at 485 nm and 0.655 at 625 nm (1.00-cmcells). Calculate the pH of the solution and Ka for the weak acid. (e) What would be the absorbance of a solution at 485 and 625 nm (1.50-cm cells) that was 2.00 10-4 M in the indicator and was buffered to a pH of 6.000?The equilibrium constant for the conjugate acid-base pair HIn+H2OH3O++In is 8.00 10-5. From the additional information in the following table, (a) calculate the absorbance at 430 nmand 600 nm for the following indicator concentrations: 3.00 10-4M,2.00 10-4M, 1.00 10-4M, 0.500 10-4 M, and 0.250 10-4M. (b) plot absorbance as a function of indicator concentration.A 0.1 M solution of acid was used to titrate 10 ml of 0.1 M solution of alkali and thefollowing volumes of acid were recorded: 9.88 10.18 10.23 10.39 10.21 Calculate the 95% confidence limits of the mean and use them to decide whetherthere is any evidence of systematic error.
- For Indirect Iodometric Analysis of Copper... ~0.0896g KIO3 necessary to consume 350mL of 0.1 M Na2S2O3, Na2S2O3 is stored in an amber glass bottle until ready for use. Primary Standard KIO3 has 2g of KI, 50mL of DI water, and 10 mL of 1.0M HCl is added then immediately titrated with Na2S2O3 until medium yellow or straw... then 5mL of starch indicator is added and titrated again until blue black color turns clear. Unknown CuO use 1.2G of Unknown, 20mL of HNO3 heated until sample dissolved, 25 mL of DI water added and boiled until clear light blue color, after cooling 1:1 NH3 added (~34.47 mL of NH4OH reagent) until permanent deep blue color amine complex, 2g of NH4HF2 added and swirled until dissolved, 3 g of KI is added then titrated immediately with Na2S2O3 until brown color of iodide is nearly gone (brown milk color), 2 g of KSCN and 3 mL of starch indicator is then added with titration continuing until disappearance of new blue black color. 1. Na2CO3 is often added to thiosulfate…For Indirect Iodometric Analysis of Copper... ~0.0896g KIO3 necessary to consume 350mL of 0.1 M Na2S2O3, Na2S2O3 is stored in an amber glass bottle until ready for use. Primary Standard KIO3 has 2g of KI, 50mL of DI water, and 10 mL of 1.0M HCl is added then immediately titrated with Na2S2O3 until medium yellow or straw... then 5mL of starch indicator is added and titrated again until blue black color turns clear. Unknown CuO use 1.2G of Unknown, 20mL of HNO3 heated until sample dissolved, 25 mL of DI water added and boiled until clear light blue color, after cooling 1:1 NH3 added (~34.47 mL of NH4OH reagent) until permanent deep blue color amine complex, 2g of NH4HF2 added and swirled until dissolved, 3 g of KI is added then titrated immediately with Na2S2O3 until brown color of iodide is nearly gone (brown milk color), 2 g of KSCN and 3 mL of starch indicator is then added with titration continuing until disappearance of new blue black color. 4. Why is the starch indicator solution…The absorbance of an iron thiocyanate solution containing 0.00500 mg Fe/mL was reported as 0.4900 at 540nm. Calculate the specific absorptivity including units of iron thiocyanate on the assumption that a 0.100 cm cuvette was used.
- If pH (± 0.5 pH units) strips were used to determine the equivalence point at pH=8.5. What is the relative error (%) in pH at the equivalence point? I assumed it should be 0.5/8.5 * 100 = 5.88, rounded to 5.9 for sig figs. but I got the answer wrong. Please help and let me know how to solve this.0 The acid-base indicator HIn undergoes the following reaction in dilute aqueous solution:HIn color1 m H1 1 In2color2The following absorbance data were obtained for a 5.00 3 1024M solution of HIn in 0.1 M NaOH and0.1 M HCl. Measurements were made at wavelengths of 485 nm and 625 nm with 1.00-cm cells.0.1 M NaOH A485 5 0.075 A625 5 0.9040.1 M HCl A485 5 0.487 A625 5 0.181In the NaOH solution, essentially all of the indicator is present as In2; in the acidic solution, it is essentiallyall in the form of HIn.(a) Calculate molar absorptivities for In2 and HIn at 485 and 625 nm.(b) Calculate the acid dissociation constant for the indicator if a pH 5.00 buffer containing a small amountof the indicator exhibits an absorbance of 0.567 at 485 nm and 0.395 at 625 nm (1.00-cm cells).(c) What is the pH of a solution containing a small amount of the indicator that exhibits an absorbance of0.492 at 485 nm and 0.245 at 635 nm (1.00-cm cells)?(d) A 25.00-mL aliquot of a solution of purified weak…Find the initial solution using Vogel's approximation method. It is necessary to fill all the cells, in the case of the blank cell it is neccessarry to select dashed line ("--"). O1 O2 O3 O4 Supply D1 18 8 13 14 200 Answer30025015010050--- Answer30025015010050--- Answer30025015010050--- Answer30025015010050--- D2 10 10 16 14 200 Answer30025015010050--- Answer30025015010050--- Answer30025015010050--- Answer30025015010050--- D3 16 15 11 14 300 Answer30025015010050--- Answer30025015010050--- Answer30025015010050--- Answer30025015010050--- Demand 50
- A CaCO3 solution that will be used to standardize EDTA was prepared by dissolving 3.1251 g of solid CaCO3 in 100 mL dilute HCl. A 20.00 mL aliquot was taken for titration with EDTA consuming 32.00 mL of the titrant to reach the endpoint. Express the concentration of EDTA in molarity and in CaCO3 titer. MM Na2H2Y2•2H2O = 372.24 g/mole; MM CaCO3 = 100.09 g/moleA CaCO3 solution that will be used to standardize EDTA was prepared by dissolving 3.1251 g of solid CaCO3 in 100 mL dilute HCl. A 20.00 mL aliquot was taken for titration with EDTA consuming 32.00 mL of the titrant to reach the endpoint. Express the concentration of EDTA in molarity and in CaCO3 titer.MM Na2H2Y2•2H2O = 372.24 g/mole; MM CaCO3 = 100.09 g/moleCompare and contrast the amount of caffeine obtained from the single vs multipleextractions with Red Bull and Coca-Cola beverages. Explain why this has occured. Single Extractions: TEST SOLUTION NAME ABSORBANCE READING [CAFFEINE]mg/100ml 5ml Red Bull 2.412 57.9mg/100ml 5ml Coca-Cola 0.716 16.2mg/100ml Multiple Extractions: TEST SOLUTION NAME ABSORBANCE READING [CAFFEINE]mg/100ml 10ml Red Bull 3.253 78.453mg/100ml 10ml Coca-Cola 0.785 17.844mg/100ml
