A 20-mL solution of 0.0659 M AgNO3 was titrated with 20 mL of 0.0509 M NaCl (Ksp = 1.8x10-10). Calculate the pAg of the analyte solution. Express your answer with 3 decimal places.
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- A 0.9092g sample of wheat flour was analyzed by Kjeldahl method. The ammonia formed was distilled into 50.00 mL of 0.05063 M HCl; a 7.46 mL back titration with 0.04917 M NaOH was required. Calculate the % protein in the flour. Input the value only. Express in 2 SF.10-27. A 10.231-g sample of window cleaner containing ammonia was diluted with 39.466 g of water. Then 4.373 g of solution were titrated with 14.22 mL of 0.106 3 M HCl to reach a bromocresol green end point. (a) What fraction of the 10.231-g sample of window cleaner is contained in the 4.373 g that were analyzed? (b) How many grams of NH3 (FM 17.031) were in the 4.373-g sample? (c) Find the weight percent of NH3 in the cleaner.You want to determine the concentration of NaCl in a water. In preliminary analysis, it appears to be saline water with a NaCl content of approximately 4%. Should you dilute the sample if your titrant has a concentration of 0.1 M and you want to spend 25 mL of sample?
- Bay Water Titration The concentration of Cl- in ocean water is about 500-600 mM. The baywater is diluted by a factor of 12.5 using a 20.00 mL volumetric pipet and a 250 mL volumetric flask. Using a 15.00 mL volumetric pipet, 15.00 mL is transferred into three clean Erlenmeyer flasks. 10 mL of 1% dextrin solution, 20 mL of de-ionized water, and 3-4 drops of indicator are added and titrated each with the AgNO3 solution. From the procedure, find: Dilution factor Volume of diluted bay water Then calculate: The [Cl-] of diluted bay water The [Cl-] of bay water Consider this information: [AgNO3] = 0.04043177 trial # Veq 1 13.91 2 13.73 3 13.9 4 13.86 5 13.87 6 13.84 average 13.85167 [Cl-]Bay = ([AgNO3](Veq(ave))/Vdil bay water pipeted) (Vflask/Vbay water)Example You determine the acetic acid content in vinegar by titrating with a standard (known concentration) solution of NaOH to a phenolphthalein endpoint. 5.023± 0.003 g sample of vinegar is weighed. The NaOH must be standardized by KHP and 3 such titrations gave molarities of 0.1167, 0.1163 and 0.1164 M. A volume of 36.78 ± 0.03mL NaOH is used to titrate the sample. What is the percent acetic acid in the vinegar and what is the overall uncertainty?Answer: 5.12% ± 0.01%1. A sample of pure sodium oxalate, Na2C2O4 , weighing 0.2856 g is dissolved in water, sulfuric acid is added, and the solution titrated at 70°C , requiring 45.12 mL of a KMNO4 solution. The end point is overrun and back-titration is carried out with 1.74 mL of a 0.1032 N solution of oxalic acid. Calculate the normality of the KMNO4 solution. Hint: A redox reaction is involed between oxalate and permanganate. Look at the changes in oxidation state by determining the balanced redox reaction: 5C20,2 + + 2MN2+ 2 Mn04 + 10H* → 10CO2 + 8H20
- 2+ as (6) A sample of an ore was analyzed for Cu* follows. A 1.25 g sample of the ore was dissolved in acid and diluted to volume in a 250 mL volumetric flask. A 20 mL portion of the resulting solution was transferred by pipet to a 50 ml volumetric flask and diluted to volume. An analysis showed that the concentration of 2+ Cu* in the final solution was 4.62 ppm. What is the weight percent of Cu in the original ore?Direct titration: 10.90 ± 0.02 mL of an unknown KHP solution required 11.72 ± 0.02 mL of 0.1090 ± 0.0006 M NaOH to reach the end point. What is the concentration of KHP in the unknown solution reported with absolute uncertainty? If the unknown solution was prepared by dissolving 5.128 g of the unknown KHP (204.22 g/mol) in 100.00 mL of water, what is the weight percent of KHP in the sample?A student performed a titration to determine the exact concentration of NaOH(aq). The titration was performed against a standard 0.1000 M HCl(aq) solution. Phenolphthalein indicator was used. The following end-point volumes were recorded by the student in units mL: Trial Number Volume 1 25.06 2 25.15 3 25.02 4 25.17 5 25.07 Calculate the 95% confidence limit. Assume that there is no outlier. Provide your answer to the correct number of decimal places, without units, and without the ± sign. Use the T-table shown below.
- Molarity of Na2S2O3 (stock solution). 0.100 M Volume of diluted bleach titrated (taken from the volumetric flask). 10ml Initial Burette Reading. 14.80 Final Burette Reading 25.60 Volume of Na2S2O3 used in the titration (mL). 10.80 Moles of Na2S2O3 used in the titration. 1.71 Moles of S2O3–2 used. 1.66 Moles of I2 initially present in titration mixture 0.83 Moles of NaOCl in the diluted bleach titrated. 0.415 Volume of bleach titrated. 1ml Mass of commercial bleaching solution titrated ?Molarity of Na2S2O3 (stock solution). 0.100 M Volume of diluted bleach titrated (taken from the volumetric flask). 10ml Initial Burette Reading. 14.80 Final Burette Reading 25.60 Volume of Na2S2O3 used in the titration (mL). 10.80 Moles of Na2S2O3 used in the titration. 1.71 Moles of S2O3–2 used. 1.66 Moles of I2 initially present in titration mixture 0.83 Moles of NaOCl in the diluted bleach titrated. 0.415 Moles of OCl– in the diluted bleach titrated. ?Molarity of Na2S2O3 (stock solution). 0.100 M Volume of diluted bleach titrated (taken from the volumetric flask). 10ml Initial Burette Reading. 14.80 Final Burette Reading 25.60 Volume of Na2S2O3 used in the titration (mL). 10.80 Moles of Na2S2O3 used in the titration. 1.71 Moles of S2O3–2 used. 1.66 Moles of I2 initially present in titration mixture 0.83 Moles of NaOCl in the diluted bleach titrated. 0.415 Volume of bleach titrated. 1ml Mass of NaOCl in commercial bleach titrated. ?