A 1.0 g sample containing urea was dissolved in water and diluted in a 100.00 mL volumetric flask. 20.00 mL aliquot was mixed with 30.00 mL water and subjected to digestion with concentrated H2SO4. The ammonia produced is collected in 30.00 mL of 0.0500 M HCI and the excess acid is back titrated with 0.0500 M NAOH, requiring 6.80 mL. Calculate the weight percent urea [(NH2)2CO, FW=60.06 g/mol] in the sample.

A 1.0 g sample containing urea was dissolved in water and diluted in a 100.00 mL volumetric flask. 20.00 mL aliquot was mixed with 30.00 mL water and subjected to digestion with concentrated H2SO4. The ammonia produced is collected in 30.00 mL of 0.0500 M HCI and the excess acid is back titrated with 0.0500 M NAOH, requiring 6.80 mL. Calculate the weight percent urea [(NH2)2CO, FW=60.06 g/mol] in the sample.

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter15: Complex Acid/base Systems

Section: Chapter Questions

Problem 15.29QAP

See similar textbooks

Similar questions

Fifty cm3 of 1.000 M nitrous acid is titrated with 0.850 M NaOH. What is the pH of the solution (a) before any NaOH is added? (b) at half-neutralization? (c) at the equivalence point? (d) when 0.10 mL less than the volume of NaOH to reach the equivalence point is added? (e) when 0.10 mL more than the volume of NaOH to reach the equivalence point is added? (f) Use your data to construct a plot similar to that shown in Figure 14.10 (pH versus volume NaOH added).
Consider the following two acids: pKa1 = 2.98; pKa2 = 13.40 HO2CCH2CH2CH2CH2CO2H Adipic acid pKa1 = 4.41; pKa2 = 5.28 In two separate experiments the pH was measured during the titration of 5.00 mmol of each acid with 0.200 M NaOH. Each experiment showed only one stoichiometric point when the data were plotted. In one experiment the stoichiometric point was at 25.00 mL added NaOH, and in the other experiment the stoichiometric point was at 50.00 mL NaOH. Explain these results.
Lactic acid is a common by-product of cellular respiration and is often said to cause the burn associated with strenuous activity. A 25.0-mL sample of 0.100 M lactic acid (HC3H5O3, pKa = 3.86) is titrated with 0.100 M NaOH solution. Calculate the pH after the addition of 0.0 mL, 4.0 mL, 8.0 mL, 12.5 mL, 20.0 mL, 24.0 mL, 24.5 mL, 24.9 mL, 25.0 mL, 25.1 mL, 26.0 mL, 28.0 mL, and 30.0 mL of the NaOH. Plot the results of your calculations as pH versus milliliters of NaOH added.
Without doing detailed calculations, sketch the curve for the titration of 30.0 mL of 0.10 M NaOH with 0.10 M HCl. Indicate the approximate pH at the beginning of the titration and at the equivalence point. What is the total solution volume at the equivalence point?
Malonic acid (HO2CCH2CO2H) is a diprotic acid. In the titration of malonic acid w ith NaOH, stoichiometric points occur at pH = 3.9 and 8.8. A 25.00-mL sample of malonic acid of unknown concentration is titrated with 0.0984 M NaOH, requiring 31.50 mL of the NaOH solution to reach the phenolphthalein end point. Calculate the concentration of the initial malonic acid solution. (Sec Exercise 113.)
Without doing detailed calculations, sketch the curve for the titration of 50 mL of 0.050 M pyridine, C6H5N (a weak base), with 0.10 M HCl. Indicate the approximate pH at the beginning of the titration and at the equivalence point. What is the total solution volume at the equivalence point?
Calculate the volume of 0.225-M NaOH required to titrate to the equivalence point for each of these samples. (a) 20.0 mL of 0.315-M HBr (b) 30.0 mL of 0.250-M HClO4 (c) 6.00 g of concentrated acetic acid, CH3COOH, which is 99.7% pure
The Merck Index indicates that 10 mg of guanidine,CH5N3, may be administered for each kilogram of bodymass in the treatment of myasthenia gravis. The nitrogenin a 4-tablet sample that had a total mass of 7.50 g wasconverted to ammonia by a Kjeldahl digestion, followedby distillation into 100.0 mL of 0.1750 M HCl. Theanalysis was completed by titrating the excess acid with11.37 mL of 0.1080 M NaOH. How many of thesetablets represent a proper dose for a patient who weighs(a) 100 lb, (b) 150 lb, and (c) 275 lb?
Consider the titration of 20 mL of 0.10 M base with 0.20 M HCl. The analyte is dibasic withpKb1 = 5.00 and pKb2 = 10.00. Calculate the pH at the following volumes of HCl0 mL, 5 mL, 10 mL, 15 mL, 20 mL
In order to determine the concentration of benzylamine (pKa = 9.33) in a urine sample, acid-base titration was performed to 100.0 mL of urine. 20.0 mL of 0.01 mM HCl was used to reach the equivalence point. What is the molarity (in M) of benzylamine? The structure of benzylamine is 2 M 0.000000002 M 0.000002 M 0.002 M 10.0 mL of 0.02 M carbonic acid (H2CO3) is titrated with 0.01 M NaOH. What is the pH of the carbonic acid solution before titration? For carbonic acid, Ka1= 4.2 x 10-7, Ka2=4.3 x 10-11.
Titration of 50.00 mL of 0.04715 M Na2C2O4 required 39.25 mL of a potassium permanganate solution. 2MnO4- + 5C2O4-2 +16H+ -> 2Mn2+ + 10CO2(g) + 8H2O Calculate the molar concentration of the KMnO4 solution. Pls show step by step. What type of molecules will interfere in this analysis ?
A 0.500 g of sample containing Na2CO3 and inert matter is analysed by adding a slight excess of 50.0 mL of 0.100 M HCl, boiling to remove CO2, and then back titrating the excess acid with 0.100 M NaOH. If 5.6 mL NaOH is required for the back titration, calculate the percent Na2CO3 in the sample