Fundamentals Of Analytical Chemistry
9th Edition
ISBN: 9781285640686
Author: Skoog
Publisher: Cengage
expand_more
expand_more
format_list_bulleted
Question
Please don't use Ai solution
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by stepSolved in 2 steps with 3 images
Knowledge Booster
Similar questions
- You want to make a solution of the primary standard sodium oxalate. It has a molecular mass of 133.999 g/mol (no error). Your analytical balance has an error of 0.2 mg. You add some weighing paper to the balance and it reads a mass of 0.23626 g. You then add sodium oxalate onto the weighing paper until the mass registers as 2.8539 g. You carefully transfer this to a 250.0 ml volumetric flask with an error of 0.1 ml and dissolve the sodium oxalate in a sufficient volume of water. What is the concentration of the oxalic acid (in M) and what is the absolute error (also in M).arrow_forwardYou were assigned to assay a product sample of milk of magnesia. A 0.600-g sample was reacted with 25.00 mL 0.10590 N H2SO4 . The excess unreacted acid in the solution required 13.00 mL of 0.09500 N NaOH when titrated to reach the methyl red end point. Determine the dosage strength of the product in terms of % Mg(OH)2 content. Type your answer in 2 decimal places, numbers only.arrow_forwardI need help calculating the values of the molarity of KMnO4 solution and the averages please. reaction equation --> 2 MnO4− + 5 C2O42− + 16 H+ → 2 Mn2+ + 10 CO2 + 8 H2Oarrow_forward
- Which among these is the most important piece of data in a gravimetric analysis? Mass of analyte Mass of weighing form Mass of crucible Mass of precipitating agentarrow_forwardExample 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%arrow_forwardA mixture has thre components. The percentage of sand was determined to be 26.04%, the percentage of benzoic acid was determied to be 25.92%, and the percentage of salt was determined to be 35.97%. Calculate the total precent recovery.arrow_forward
- 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.arrow_forwardA 0.1093-g sample of impure Na2CO3 was analyzed by the Volhard Method. After adding 50.00 mL of 0.06911 M AGNO3, the sample was back titrated with 0.05781 M KSCN, requiring 27.36 mL to reach the end point. Report the purity of Na2CO3 sample.arrow_forwardA 0.1093-g sample of impure Na2CO3( Molecular weight 106) was analyzed by the Volhard method. After adding 50.00 mL of 0.06911 M AGN03 (Molecular weight 169.87), the sample was back titrated with 0.05781 M KSCN, requiring 27.36 mL to reach the end point. Report the purity of the Na2CO3 sample.arrow_forward
- Silver nitrate can be standardized using primary standard KCl. A dried sample of analytical grade KCl of mass 0.918 g was dissolved and diluted to 250.0 mL. Repeat 10.00-mL aliquots of the potassium chloride solution were titrated with the silver nitrated solution. The mean corrected titration volume was 8.98 mL Calculate the molarity (M) of the silver nitrate solution.arrow_forwardPls help ASAP. Pls show all work and calculations.arrow_forwardYou took 25.00 mL of unknown water solution and titrated it with EDTA. The EDTA solution molarity was 0.01 M, and it took 16.50 mL to reach the end point. With the blank it took only 0.98 mL to reach end point. 15.52 mL of EDTA solution were used to titrate hardness that actually came from the unknown 1.552×10−4 moles of EDTA reacted with hardness-causing ions from the unknown sample 1.552×10−4 moles of hardness-causing ions were present in the unknown sample A) Assuming that the total hardness of water is due to CaCO3, how many grams CaCO3 does it correspond to? B) What is the total hardness of the unknown water in ppm CaCO3?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you