I.
Encircle the best answer. 1. Double indicator titration is used to determine amounts of NaOH, NaHCO 3 and Na 2CO3 in a solution. Phenolphthalein and methyl orange are the indicators used. Which of the following is/are TRUE regarding double indicator titration? I. Volumes of indicators used can determine the concentration of the 3 chemicals in solution. II. Only methyl orange is the t he necessary indicator to determine the concentration of NaHCO 3. III. In the continuous titration of a sample containing a mixture of NaOH and Na 2CO3, the volume of HCl used until the endpoint of phenolphthalein is greater than the volume of HCl used unti l the endpoint of methyl orange. A. B. C. D.
I only III only II and III I, II and III
2. Which of the following methods do not promote the formation of solid precipitate? A. Decreasing the temperature of the solution. B. Slow addition of precipitating agent with vigorous stirring. C. Diluting the solution. D. Increasing the solubility of the solution. 3. What is the ionic strength s trength of a 0.0250 M Na 3PO4 solution? A. 0.125 M B. 0.150 M C. 0.300 M D. 0.350 M 4. A 0.5000 g tablet containing CaCO 3 was initially titrated with 20.00 mL of 0.1250 M HCl. Afterwards, the excess acid was titrated using 2.80 mL of 0.1000 M NaOH. What is the % of CaCO 3 in the tablet? A. 19.42% B. 22.22% C. 44.44% D. 88.88% 5. A chemistry laboratory received 15 mg of soil sample from the geology laboratory. The sample was divided into three 5 mg samples and was analyzed. What What is the classification of the analytes according to amount? A. Macro B. Semimicro C. Micro D. Ultramicro 6. What type of water is held as condensed phase in capillaries of colloidal solids? A. Essential water B. Adsorbed water C. Sorbed water D. Occluded water
CHEM 28 (1st sem, AY 13-14)
UP 49ers, UP ACME | Page of
7. A wheat sample containing 19.60% Ca as received became completely dried after oven-drying. What is the moisture content of the sample if it contains 27.50% Ca after it was dried? A. 28.73% B. 71.27% C. 5.39% D. 94.61% 8. Which of the following is NOT TRUE about gravimetric analysis? A. Reaction should be complete and quantitative. B. In precipitation gravimetry, precipitate with high solubility can be used. C. All substances weighed must be pure and must have definite chemical composition. D. In volatilization gravimetry, the analyte is converted to a gas of known chemical composition. 9. Which of the following is TRUE about precipitate formation? A. Particle size is directly proportional to relative supersaturation. B. Rapid addition of precipitating agent with good stirring will lower the concentration of the solute at any instant. C. If crystal growth predominates, larger number of small particles will be obtained. D. If crystal growth predominates, fewer particles of large particle size will be obtained. 10. The process of converting a colloidal suspension into a filterable solid is called A. Coagulation B. Peptization C. Coprecipitation D. Nucleation 11. Which of the following is NOT a good characteristic of a primary standard? A. Reasonable solubility in titration medium B. Must have small molecular mass for easier computation C. Stable in the atmosphere D. Readily available at reasonable cost 12. A 0.4000 g sample composed of CaO and other impurities is mixed with 14.00 mL of 0.5000 M HCl. The solution was then titrated with 4.80 mL 0.3000 M NaOH. What is the percent purity of CaO in the sample? (MM: H = 1.01, Cl = 35.45, Na = 23.00, O = 16.00, Ca = 40.08) A. 77.95% B. 49.07% C. 20.17% D. 38.98% 13. The amount of phosphorus in a 5.0 g fertilizer sample was found to be 0.047 g. The element can thus be considered to be what type of constituent? A. Major B. Minor C. Trace D. Ultratrace 14. Which of the following solutions has the lowest ionic strength? A. 0.20 M CaCO3 and 0.30 M K 2Cr2O7 B. 0.20 M CaCO3 and 0.40 M K 2Cr2O7 C. 0.30 M CaCO3 and 0.20 M K 2Cr2O7 D. 0.30 M CaCO3 and 0.40 M K 2Cr2O7
CHEM 28 (1st sem, AY 13-14)
UP 49ers, UP ACME | Page of
15. Which of the following processes will NOT hasten t he coagulation of colloids? A. Stirring the solution B. Heating the solution C. Adding an electrolyte D. Diluting the solution 16. Which of the following solutions has the lowest pH? A. 0.054 M HCl B. 0.17 M formic acid and 0.24 M sodium formate (K a = 1.8 × 10 -4) C. 1.2 M NaOH D. 0.43 M ammonia and 0.76 M ammonium chloride (K a = 5.7 × 10 -10) 17. 25.00 mL of solution are found to contain 3.45 × 10 -8 moles of fluoride ion. What is ppb F - for this solution? (MW = 19.0) A. 0.0262 ppb B. 0.655 ppb C. 1.38 ppb D. 26.2 ppb 18. 21.3 mL of 0.10 M NH 3 is mixed with 15.6 mL of 0.10 M HCl. What is the pH of the resulting solution? (K b = 1.75 × 10 -10) A. 4.32 B. 5.20 C. 8.80 D. 9.68 19. Which of the following is NOT a buffer solution? A. HF and NaF B. HNO3 and NaNO3 C. HNO2 and NaNO2 D. NaHCO3 and Na2CO3 20. What are the α-values of acetic acid with pH 5? (Ka = 1.8 × 10 -5) A. α0 = 0.643 ; α1 = 0.357 B. α0 = 0.357 ; α1 = 0.643 C. α0 = 0.847 ; α1 = 0.153 D. α0 = 0.153 ; α1 = 0.847 II.
Supply the answer being asked for. __________________1. __________________2. __________________3. __________________4. __________________5. __________________6. __________________7. __________________8. __________________9. __________________10.
Kjeldahl method is the standard method for determining _________ content. If the indicator range for a newly-discovered indicator is between 4.3 and 6.3, find the pKa of this indicator. Analytical method used to determine amount of analyte from amount of product formed. The type of analysis used to determine amount of certain selected constituents in a sample. The material taken from the lot for analysis or archiving. The chemical method used to determine moisture in a sample by non-aqueous titration. A type of coprecipitation wherein a contaminant compound is trapped within a pocket formed during rapid crystal growth. It is a process by which a precipitate is formed by slow generation of a precipitating agent homogeneously throughout a solution. The _________ of a solution is the mass of a pure substance that is chemically equivalent to 1 mL of the solution. An ultrapure compound that serves as the reference material for a titrimetric method of analysis.
CHEM 28 (1st sem, AY 13-14)
UP 49ers, UP ACME | Page of
III. 1. While searching through a storage room, a curious scientist ran through a reagent bottle which was labeled “5000 ppm X2-, 7500 ppm Y -“. This scientist used gravimetric analysis to determine the identities of X 2- and Y-. Separate 50.00 mL solutions were poured onto each beaker, and each beaker was treated with excess Mg2+ and Pb 2+, respectively. In the beaker treated with Pb 2+, after filtering the solution, the recorded weight was determined to be 3.812 g. However, the scientist forgot to oven dry the precipitate. After oven-drying, the weight was determined to be 3.773 g. In the beaker treated with Mg 2+, the recorded weight, after oven drying, was determined to be 0.6430 g. It was then determined that only MgY 2 was present using one of your earlier Chem 16 experiments. The following is a list of possible identities of X 2- and Y-, with their corresponding molecular weights:
ClBrIOH-
A. B. C. D.
MW (g/mol) 35.453 79.904 126.9045 17.0073 62.0049 59.0439
MW (g/mol) 60.0089 96.0626 80.0632 115.9937 32.065 112.1282 88.019
S2-
Determine the amount of X2- and Y- (in grams) in a 50 mL solution. Calculate the % moisture of the filtered lead precipitates. Determine the identity of Y-. Determine the identity of X2-.
2. (Quantitative Chemical Analysis, Harris, 7 th ed. 27-20) A 1.475 g sample containing NH 4Cl (FM = 53.492), K2CO3 (FM = 138.21), and inert ingredients was dissolved to give 0.100 L of solution. A 25.0 mL aliquot was acidified and treated with excess sodium tetraphenylborate, Na+B completely:
-
, to precipitate K + and
ions
FM = 358.33 FM = 337.27 The resulting precipitate amounted to 0.617 g. A fresh 50.0 mL aliquot of the original solution was made alkaline and heated to drive off all the NH 3:
It was then acidified and treated with sodium tetraphenylborate to give 0.554 g of precipitate. A. B. C. D. E.
Find the amount of K + ions (in mol) in the 25.0 mL and 50.0 mL aliquots. Find the mass of K 2CO3 found in the sample. Find the amount of ions (in mol) in the 25.0 mL aliquot. Find the mass of NH 4Cl found in the sample. Find the weight percent of K 2CO3 and NH 4Cl in the sample.
3. Two Chem 28 students, Keena and Lorainne, were asked to determine the contents of two different 200 mL soda ash samples by titration with 0.1600 M HCl as the titrant.
CHEM 28 (1st sem, AY 13-14)
UP 49ers, UP ACME | Page of
Keena, assigned with Sample A, titrated a 100 mL aliquot of Sample A with 113.76 mL titrant until the phenolphthalein endpoint was reached. She continued titrating until the methyl orange endpoint, and 48.13 mL titrant was used. Lorainne, assigned with Sample B, titrated a 50 mL aliquot of Sample B with 5 mL titrant until the phenolphthalein endpoint was reached. She titrated another 50 mL aliquot with 57.81 mL titrant until methyl orange endpoint was reached. A. B. C. D. E.
What component/s is/are present in Sample A? What is/are the concentration/s of the component/s of Sample A (in molarity)? What component/s is/are present in Sample B? What is/are the concentration/s of the component/s of Sample B (in molarity)? After the experiment, the two students decided to mix their individual remaining 100 mL aliquots. By continuous titration, how much titrant is needed to reach the phenolphthalein and methyl orange endpoints, respectively?
4. A 1.15 g sample containing only NaBr, CaBr 2 and Na2CO3 yielded 1.08 g KBr and 0.682 g Na 2SO4 in separate precipitations. Find the percent composition of each compound in the original sample. 5. In an acid-base titration, 25.00 mL of 0.1000 M oxalic acid was titrated with 0.1000 M NaOH ( 10-2 ; = 5.4 × 10 -5). Calculate the following: A. B. C. D. E. F.
= 5.6 ×
Initial pH of the solution pH before the 1 st equivalence point (VNaOH = 5.00 mL) pH on the 1 st equivalence point pH between the 1st equivalence point and the 2 nd equivalence point (VNaOH = 28.00 mL) pH on the 2 nd equivalence point pH beyond the 2nd equivalence point (VNaOH = 51.00 mL)
CHEM 28 (1st sem, AY 13-14)
UP 49ers, UP ACME | Page of
