Lab F report

Following the monograph procedure, a 724-mg of aspirin (MW=180 g/mol) dissolved in 18.5 ml of cold neutralized alcohol. This solution was then initially titrated with 0.101 N sodium hydroxide solution, then later neutralized with 0.104 sulfuric acid.   1. What is the total volume (mL) of excess titrant used in the titration? a. 56.400 mL b. 52.000 mL c. 44.600 mL   2. What is the milliequivalent weight consumed by the acidic titrant? a. 1.5392 g-meq b. 2.0907 g-meq c. 1.4948 g-meq d. 2.1528 g-meq   3. What is the milliequivalent weight consumed by the basic titrant? a. 5.8656 g-meq b. 1.5392 g-meq c. 5.6964 g-meq d. 1.4948 g-meq   4. What is the difference of milliequivalent weight consumed in the reaction? a. -4.1572 g-meq b. 4.3708 g-meq c. 0.5515 g-meq d. 4.1572 g-meq   5. What is the percentage purity of the sample?   6. What is the titer value (mg) based from the specific type of titration?

1. A 0.1250-g sample of pure (100%) of K2Cr2O7 was dissolve in 250-mL volumetric flask, and dilute to volume, producing a concentration of 1.700 × 10-3 M. The resulting solution is an example of a standard solution prepared from a primary standard. Select one: a. True b. False   2. Which of the following is not an example of a standard solution? a. 1000 ppm Lead Nitrate solution b. 1.000g pure sodium carbonate dissolved in 1L of solution c. 3:4 nitric acid: hydrochloric acid solution d. NaOH solution standardized using KHP e. 0.104 M NaOH

Data for titration of 15.00 mL of vinegar with approximately 1.0 M NaOH. Note you must use the exact concentration of the standard NaOH solution to calculate the moles of NaOH. Quantity Еxample Trial 1 Trial 2 Trial 3 M NaOH (exact concentration) 0.992 M 0.953 M 0.953 M 0.953 M V initial buret reading = V, 0.20 mL 1.20 mL 0.52 mL 0.15 mL V final buret reading = V, 12.90 mL 14.22 mL 13.71 mL 13.31 mL %3D Vep = VNAOH added = V; - V, 12.70 mL 13.02 mL 13.19 mL 13.16 mL Vep = VNAOH in L 0.01270 L 0.01302 L 0.01319 L 0.01316 L moles NaOH = 0.0126 mol 0.0124 mol 0.0126 mol 0.0125 mol MNAOH X VNAOH moles AA = 0.0126 mol 0.0124 mol 0.0126 mol 0.0125 mol moles NaOH V sample = V acid 0.0150 L 0.0150 L 0.0150 L 0.0150 L (15.00 mL) (15.00 mL) (15.00 mL) Actual molarity 0.840 M 0.827 M 0.840 M 0.833 M of AA Average molarity for 3 trials => of AA 0.833 M 1. Report the acetic acid (AA) concentration of vinegar in units of molarity (M). 2. Use your average AA concentration to calculate the mass…

M 23 (H) Inorganic Analytical Chemistry pard My courses CHEM 23 (H) FS-AY:20-21 Week 4-5: Quantitative Analysis by Gravimetry and Titration Quiz 4 Qu 10 The 500.0-mg sample of impure Na2 CO3 (FM: 105.96) required 22.00 mL of the HCI standard solution (the obtained concentration of HCI from the previous question, problem 5: Titrimetry short answer type) for complete neutralization. ut of 1. Give the balanced chemical equation between the sample, Na2CO3 and the standard, HCI. 2. What is the mass (g) of Na, CO3 in the sample? 3. Calculate the % (w/w) purity of a 500.0-mg sample of impure Na2 CO3 Note: Type your solution here or upload its pic in Jpeg or pdf format. Another option is to send it to me via email (subject: Fam name-Quiz 4) or messenger, privately. Do not forget to box the final answer and write your name. В I 1.

Following the monograph procedure, determine the weight in grams of sodium carbonate (MW-106 g/mol) used to standardize a 0.987 N sulfuric acid solution. 1. Consider that the burette was completely filled to the 0 mL mark before titrating. What is the volume of titrant consumed based from the image below? A. 22.9 mL B. 21.2 mL C. 21.3 mL D. 22.7 mL 2. What is the unknown weight (grams) in the problem? Your Answer:

Chemistry H,N. CH, H,N. HO, но Phenylalanine (1) Tyrosine (2) H,C HN CH, H,C `CH, H,C но Triethylamine (4) N-acetyl-OMe-Tyrosine (3) All four compounds are dissolved in ethyl acetate. If you started by extracting the solution with 1 M HCI, which of the following compounds would end up in the aqueous layer? 4 12 Instead, you decide to start by extracting the ethyl acetate solution with 5% NaHCO3 (sodium bicarbonate). What will remain in the organic layer? 1 2 4

Only typde Solution.

Answer the following: Maria was given a capsule of multivitamins and she was asked to determine the % by mass (w/w) of ascorbic acid present in the sample. The student analyzed 1.032 g sample using volumetric titration. Use the table as a reference of the data to be used for your solution.   Given Choices: A. 66.0 % B. 54.6%  C. 65.6% D. 7.30%

Preparation and Standardization of KMnO4 solution Experimental data Complete the table below. Trial 1 0.2001 g Trial 2 0.2065 g Trial 3 Weight of sodium oxalate (Na2C2O4, MM= 134 g/mol) Titration data 0.2050 g Final reading Initial reading 29.86 mL 0.00 mL 30.66 mL 30.52 mL 0.00 mL 0.00 mL Total vol. of KMNO4 used Computed Molarity of KMNO4 solution Mean Molarity Computed Normality of KMNO4 Mean Normality of KMNO4 solution Reaction Involved: Calculations:

Sample A B C D Identity Rationalization

A. Calculate the solubility of PbI2  (Ksp = 1.5 x 10-8) in moles per liter. Ignore any acid–base properties. B. Calculate the solubility of CdCO3 (Ksp = 5.1 x 10-12) in moles per liter. Ignore any acid–base properties. C. Calculate the solubility of Sr3(PO4)2 (Ksp = 1.5 x 10-31) in moles per liter. Ignore any acid–base properties.

9. wo samples of the same soil is subjected to pH analysis using a pH meter, the first is analysed using 1:1 soil:water ratio and the other with 1:1 soil:KCl ratio, what would be the most probable result?        A. Soil analused using 1:1 soil:KCl ratio will register a higher pH compared to that of soil analysed using 1:1 soil water ratio       B. Soil analysed using 1:1 soil:KCl will register a lower pH compared to that of soil analysed using 1:1 soil:water ratio       C. Soil analysed using 1:1 soil:water ratio will register a lower pH compared to that of soil analysed using 1:1 soil:KCl ratio       D. Soil analysed using 1:1 soil:water ratio will have the same pH with that of soil analysed using 1:1 soil:KCl ratio

REPORT SHEET Saponification and Soaps A. Saponification: Preparation of Soap Describe the appearance of the soap you prepared. A longht yelow Cyoih line)mirture upon (yoik heating B. Properties of Soaps and Detergents Tests Lab Soap Commercial Soap Detergent Bayic neutal YeS Basic Yes NO 1. pH Yes 2. Foam NO NO 3. Oil 4.1% CaCl2 コ PP+ Yes Ppt. 1% MgCl2 1% FeCl3 Questions and Problems Q1 Which of the solutions in Part B were basic? Q2 How do soaps made from vegetable oils differ from soaps made from animal fat?

A 1.5000-g sample of cereals was analyzed for nitrogen using the Kjeldahl procedure. The receiving flask contained 69 mL of 0.02 M HCl. After the ammonia was collected, the solution was titrated with 0 M NaOH, requiring 11.72 mL to reach the methyl red endpoint. Calculate and percent protein in the sample (f=5.70). Ans. in 3 SFs

A RbOH solution is titrated four (4) times against potassium hydrogen phthalate (KHP; FW=204.224) samples to the Phenolphthalein endpoint. Using the data below, determine the concentration of the RbOH solution? g of KHP Volume of Base Required 0.5373 g 42.49 mL 0.5856 g 43.88 mL 0.5790 g 48.56 mL 0.5856 g 44.60 mL (Report your answer as "mean +/- std dev") M What is the percent relative standard deviation? % What is the 99% Confidence Interval for the concentration of the solution (population mean)?

A 625 mg sample of sodium bicarbonate tablet required 7.30 mL of 0.9580N sulfuric acid solution to reach the methyl red endpoint. PRELIMINARY QUESTION: Which of the following statements is/are TRUE? I. The țitrant used is sulfuric acid solution. II. The end point is a red/pink solution. III. This is an example of direct assay titration.

5 1 X MasteringChemistry erty.edu/webapps/assessment/take/launch.jsp?course_assessment_id%=_1575667_1&course_id%3_592130_1&content_id= 39106439 18 puTube Maps M Gmail YouTube 2 Maps ¥ Question Completion Status: Which is more basic - an aqueous solution of borax or an equivalent concentration aqueous solution of lime water? [H*] (M) pH pOH [OH ] (M) 1 (1x10) 0.0 14.0 1x10-14 Stomach acid 1x10-1 1.0 13.0 1x10 13 Lemon juice Cola, vinegar 1x10-2 2.0 12.0 1x10-12 1x10-3 3.0 11.0 1x10-11 Wine Tomatoes 1x10-4 4.0 10.0 1x10-10 Black coffee 1x10-5 5.0 9.0 1x10-9 Rain Saliva Milk 1×10-6 6.0 8.0 1×10-8 1×10-7 7.0 7.0 1×10-7 Human blood 1×10-8 8.0 6.0 1x10-6 Seawater 1x10-9 9.0 5.0 1x10-5 Borax 1x10-10 10.0 4.0 1x10-4 Lime water 1x10-11 11.0 3.0 1x10-3 Household ammonia Household bleach 1x10-12 12.0 2.0 1x10-2 1x10-13 13.0 1.0 1x10-1 1x10-14 14.0 0.0 1 (1x10) pH + pOH = 14 %3D [H*][OH¯]= 1×10 14 A. borax B. household ammonia C. lime water D. the two solutions will have equivalent pH at…

Report Sheet for Experiment 13: Acid-Base Titration Trial 1 Trial 3 (if needed) Trial 2 1. Mass of KHP 0.949g 1.019g 2. Initial burette reading 6.21 mL 22.90mL 3. Final burette reading 15.81mL 33.10m 10.2mL 4. Volume of NaOH used 9.6mL Calculations 5. Moles of KHP (MM of KHP = 204.2 g/mol) (Show one set-up) 6. Moles of NAOH sol'n (Hint: mol ratio) (Show one set-up) 7. Molarity of NaOH sol'n (Hint: M = mol/L) (Show one set-up) 8. % Difference between trial 1 and 2 (Show one set-up) |Molarity trial #1 - Molarity trial #2| × 100% s 5% 9. Average Molarity of NaOH sol'n (Show set-up)

Saved CHEMY 101 SEC SEM T2 2020-21 The amount of water which should be added to 0.700 M NAOH solution to make 100 mL 0.250 M solution is 19 Multiple Choice 01:25:58 79.6 mL 35.7 mL 55.2 mL 64.3 mL raw K Prev 19 of 20 Next >

An unknown sample containing mixed alkali (NAOH, NaHCO3, or NazCO3) was analyzed using the double flask method. A 250 mg sample was dissolved in 250 mL CO2 free water. A 20.0 mL aliquot of this sample required 11.3 mL of 0.009125 M HCI solution to reach the phenolphthalein end point. Another 29.0 mL aliquot of the sample was titrated to the bromocresol green endpoint using 31.1 mL of the standard acid. How many millimoles of the components are there in the original solid sample? O 0.28 mmol Na,CO3 O 1.29 mmol NaOH, 0.97 mmol Na2CO3 O 0.103 mmol NaOH, 0.078 mmol NazCO3 O 1.29 mmol Na2CO3, 0.97 mmol NaHCO3 O Cannot be determined

An unknown sample containing mixed alkali (NAOH, NaHCO3, or NazCO3) was analyzed using the double flask method. A 250 mg sample was dissolved in 250 mL CO, free water. A 20.0 mL aliquot of this sample required 11.3 mL of 0.009125 M HCI solution to reach the phenolphthalein end point. Another 29.0 mL aliquot of the sample was titrated to the bromocresol green endpoint using 311 mL of the standard acid. What is/are the component/s of the sample? O Cannot be determined O NAHCO3 only O NAOH and Na,CO3 O NazCO3 only O NAHCO, and Na2CO3

Standard solution : 0.201 grams of acetylsalicylic acid was combined with 10 mL of 0.5 M NaOH, then transfered to a 100 mL flask and filled the rest of the way with water Then 0.500 mLs of the standard solution was transfered to a 10.00 mL flask and diluted to the 10 mL mark with 0.02 M of buffered iron chloride.  Number of moles of NaOH = 0.005 moles Since the reaction between acetylsalicylic acid and NaOH is a 1:1 reaction, the number of moles of acetylsalicylic acid used in the reaction is also 0.005 moles. the concentration from analysis question 5: acetylsalicylic acid in the 100.00 mL volumetric flask is 0.050 M. From your plot, what is the value of εb? From the Beer- Lambert law For the aspirin sample, calculate the concentration of acetylsalicylic acid present using the value of εb that you found   The concentration from analysis question 5 represents the concentration in the 10.00 mL sample that was prepared in the volumetric flask using an aliquot of the solution in the…

An unknown sample containing mixed alkali (NAOH, NaHCO3, or NazCO3) was analyzed using the double flask method. A 250 mg sample was dissolved in 250 mL CO, free water. A 20.0 mL aliquot of this sample required 11.3 mL of 0.009125 M HCI solution to reach the phenolphthalein end point. Another 29.0 mL aliquot of the sample was titrated to the bromocresol green endpoint using 31.1 mL of the standard acid. What is/are the component/s of the sample? O NazCO3 only NaHCO3 only O NAHCO3 and Na>CO3 NaOH and NazCO3 O Cannot be determined

1. Prepare 50-mL of a 50 mM H,PO, solution in a 50-ml conical tube. Molar mass of NaH2PO4 H2O used is 138 moles 2. Check the pH of this solution and record. 3. A5 M NAOH solution has been provided. 4. Add 45 µl of 5 M NaOH into the 50 mM H2PO, solution. Check the pH and record. 5. Add 205 µl of 5M NaOH into the 50 mM H2PO4 solution. Check the pH and record. 6. Add 205 µL more of 5M NaOH into the 50 mM H2PO, solution. Check the pH and record. 7. Add 45 µl of 5 M NaOH into the 50 mM H2PO, solution. Check the pH and record. Complete the following table "predicted pH." Please show all work on a separate sheet of paper and attach. Note that the pK, of H2PO, is 7.2. What would be the predicted pH's? Would they be similar to the actual pH's from procedure? After 45 µl of 5 | After 205 µl more of 5M NaOH Original After 205 µl more After 45 µl more solution M NaOH of 5M NaOH of 5 M NaOH Actual pH 4.30 6.05 6.90 11.01 11.29 Predicted pH

An unknown sample containing mixed alkali (naoh, nahco3, or na2co3) was analyzed using the double flask method. a 250 mg sample was dissolved in 250 ml co2 free water. a 20.0 ml aliquot of this sample required 11.3 ml of 0.009125 m hcl solution to reach the phenolphthalein end point. another 29.0 ml aliquot of the sample was titrated to the bromocresol green endpoint using 31.1 ml of the standard acid. how many millimoles of the components are there in the original solid sample

100 mL water sample was subjected to DO analysis using the Winkler method. The initial reading of the titrant (0.025 M Na2SO3) in the buret was 2.51 mL. The endpoint was obtained and the final reading was 7.07 mL. A blank was also titrated and 0.75 mL of the titrant was consumed. Compute for the amount of DO (mg/L) in the sample.

3. Titration: Strong Acid / Weak Base. (I warned you to be ready!) Here's the cool chart! Analyte: 20.00 mL 0.100 M Hydrazine (N₂H4) Hd A: initial Vertical (Value) Axis Title 0 B: half equivalence C: equivalence 5 10 15 20 Titrant: mL of 0.100 M HCI added 25 20.00 mL of 0.100 M hydrazine (N₂H4, pKb = 5.77) is titrated with 0.100 M HCl. a) Find the initial pH (point A) and the pH at the equivalence point (point C). b) Find the pH at the half equivalence point (point B).

O Macmillan Learning Na₂CO3 is used to standardize a solution of HC1. What is the concentration of HCl if 0.1970 g of pure Na₂CO3 dissolved in 50.00 mL of water took 29.41 mL of HCl to reach the end point in the titration? Next, 0.4242 g of an unknown sample containing Na2CO3 was dissolved in 50.00 mL of water. It took 18.92 mL of the standardized HCl solution to reach the end point in the titration of the unknown sample. What is the wt% of Na₂CO3 in the unknown? M HCI wt % Na₂CO3

MISSED THIS? Read Section 18.6 (Pages 823-826); Watch IWE 18.12. Determine the minimum concentration of the precipitating agent on the right to cause precipitation of the cation from the solution on the left. 9.0x10-2 M Cala; K2SO4 Express your answer using two significant figures. [K₂SO4]= ΜΕ ΑΣΦ Submit Request Answer Part C 1.7x103 M AgNO3; RbCl Express your answer using two significant figures. [RbCl] ΜΕ ΑΣΦ Submit Request Answer M M

5. The concentration of H₂SO3 in a saturated aqueous solution at room temperature is approximately 0.10 M. Calculate [H3O*], [HSO3], and [SO32-] in the solution: Ka1 1.0 x 10-2, Ka2 = 6.0 x 10-8 1 C e + H₂SO₂H₂O + HSO₂" HS03 0.10 M 0.10M-X O +X X 2 (H₂0°) (HSO₂) _x² (H₂SO₂) (0.10-x) Val =1₁0×10²² = x ² (0.10) 0.40 10.01=1x² 0.032= X Assume x <<<0110-x 0.10- 20.10 (040) Validity = 0.032x100, 0.10 = 0.32% <5%

Following the monograph procedure, a 724-mg of aspirin (MW+180 g/mol) dissolved in 18.5 ml of cold neutralized alcohol. This solution was then initially titrated with 0.101 N sodium hydroxide solution, then later neutralized with 0.104 sulfuric acid.   1. What is the milliequivalent weight consumed by the acidic titrant? a. 1.5392 g-meq b. 2.0907 g-meq c. 1.4948 g-meq d. 2.1528 g-meq   2. What is the milliequivalent weight consumed by the basic titrant? a. 5.8656 g-meq b. 1.5392 g-meq c. 5.6964 g-meq d. 1.4948 g-meq   3. What is the difference of milliequivalent weight consumed in the reaction? a. -4.1572 g-meq b. 4.3708 g-meq c. 0.5515 g-meq d. 4.1572 g-meq