A- STANDARD PHOSPHATE SOLUTION: Standard solution 1 4 6. Conc of PO4 (ppm) O-00 O.2 0.4 0.6 0.8 1-0 Absorbance 0-00 O0 33 0:069 O-107 0.134 0.155 B - UNKNOWN WATER SAMPLE: Unknown Water Sample Number: Absorbance reading: 0.083 Calculated Concentration (P04"): CALCULATIONS Set each problem completely and SHOW UNITS. Cancel the units when you do the up calculations. Use the correct number of significant figures in your answers. 1. Determine the ppm of Phosphate in your water sample. by us Inconcent

A- STANDARD PHOSPHATE SOLUTION: Standard solution 1 4 6. Conc of PO4 (ppm) O-00 O.2 0.4 0.6 0.8 1-0 Absorbance 0-00 O0 33 0:069 O-107 0.134 0.155 B - UNKNOWN WATER SAMPLE: Unknown Water Sample Number: Absorbance reading: 0.083 Calculated Concentration (P04"): CALCULATIONS Set each problem completely and SHOW UNITS. Cancel the units when you do the up calculations. Use the correct number of significant figures in your answers. 1. Determine the ppm of Phosphate in your water sample. by us Inconcent

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter27: Molecular Fluorescence Spectroscopy

Section: Chapter Questions

Problem 27.11QAP

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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…
Q1. Dissolved 0.273 grams of pure sodium oxalate (Na,C,O.) in distilled water and added sulfuric acid and titration the solution at 70 ° C by using 42.68 ml of KMNO, solution and has exceeded end point limits by using 1.46 ml of standard oxalic acid (H; C;O.) with 0.1024 N. Calculate the normlity of KMN0.. Note that the molecular weight of sodium oxalate (Na,C,O.) = 134 and its equivalent weight = 67
A chemist received different mixtures for analysis with the statement that they contained NaOH, NaHCO3, Na2CO3, or compatible mixtures of these substances, together with inert material. From the data given, identify the respective materials, and calculate the percentage of each component. 1.000g samples and 0.2500N HCl were used in all cases. (A) For Sample W: With PP, 24.32ml was used. A duplicate sample required 48.64ml with MO. (B). For Sample X: The addition of PP caused no color change. With MO, 38.47ml of the acid was required. (C). For Sample Y: To cause a color change in the cold with PP, 15.29ml of the acid was necessary, and an additional 33.19ml was required for complete neutralization. (D) For Sample Z: The sample was titrated with acid until the pink of PP disappeared; this process required 39.96ml. On adding an excess of the acid, boiling, and titrating back with alkali, it was found that the alkali was exactly equivalent to the excess acid added.
A stock solution of analyte is made by dissolving 34.83 mg of copper (II) acetate hexahydrate(fw = 289.73 g/mol) in 25.00 mL of water. A second stock solution of internal standard is madeby dissolving 28.43 mg of germanium (I) acetate (fw = 190.74 g/mol) into 25.00 mL of water.These solutions are used to make a series of standards for ﬂame atomic absorption analysiscalibration. The standard solutions (each 10.00 mL total volume) should have the followingconcentrations of copper: 10.00; 25.00; 50.00; 100.0; and 200.0 μM. Each calibration solutionshould also contain 50.00 μM of germanium. What is the analyte stock solution concentration? What is the internal standard stocksolution concentration?Complete this table.
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.
A 30.00-L air sample was passed through an absorption tower containing a solution of Cd+2, where H2S was retained as CdS. The mixture was acidified and treated with 10.00 mL of 0.01070 M I2. After the reaction: S-2 + I2 S(s) + 2 I- was complete, the excess iodine was titrated with 12.85 mL of 0.01344 M thiosulfate. Calculate the concentration of H2S in ppm (w/w); use 1.20 g/L for the density of the gas stream. Molar Mass: H2S = 34.082
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 accompanying data (1.00-cm cells) were obtained for the spectrophotometric titration 10.00 mL of Pd(II) with 2.44 10-4 M Nitroso R(O. W Rollins and M. M. Oldham, Anal. chem .,1971, 43, 262, DOI: 10.1021/ac60297a026). Calculate the concentration of the Pd(II) solution, given that the ligand-to-cation ratio in the colored product is 2:1
A sample of anhydrous Na2CO3 (FM = 105.989) is suspected to be contaminated with eitherNaHCO3 (FM = 84.007) or NaOH (FM = 39.997) To verify the suspicion, a 0.7483 g sample was dissolved to prepare a 50.00 mL solution, and a 10.00 mL aliquot was taken to prepare a 100 mL solution, where 25.00 mL was analyzed using single flask method. If the sample requires 27.50 mL of standard 0.0125 N HCl to reach the phenolphthalein endpoint and another 28.40 mL to reach the bromcresol green endpoint, calculate the percentagecomposition (in %w/w) of all the basic components in the sample.
A 0.3045 g of CaCO3 primary standard was dissolved using concentrated HCl, evaporated to incipient dryness and quantitatively transferred to a 250 mL volumetric flask, and diluted to mark. A 10 mL of aliquot was then transferred to an Erlenmeyer flask, together with 5mL buffer and 5 drops of EBT indicator, and was used to standardize the EDTA titrant. The solution turned blue after the addition of 24.10 mL of EDTA. A blank correction was performed and was determined to be 0.4 mL. What is the concentration of EDTA obtained (MW CaCO3 = 100.0869 g/mol)?
Six iron tablets containing FeSO4.7H2O were dissolved in 100-ml of 0.1M HNO3 with gentle heating. All of the Fe2+ is converted to Fe3+ by the strong oxidizing conditions. After the solution had cooled to room temperature , 2.5-ml of 35wt% NH4OH was added. The precipitate Fe2O3-xH2O that was filtered weighed 0.345g. Thermogravimetric analysis of the crude product showed a 10.5% weight loss . A. How many waters of hydration were in the precipitate B. How much iron is present in each tablet