Analysis of Data: Calculate the molarity f the HCI for each trial that was titrated using your volume of NaOH used. Show all work with proper units and report your answer below. Continue on the back if necessary. NaOH + HCI → H,0 + NaCI Calculate the average molarity of the HCI that was titrated. Show your work. Average molarity of HCI:
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- Which of the following steps MUST be done prior to titration of an analyte? I. Prepare a standardized solution II. Wash the burette with a standardized solution III. Fill the burette to the 50-mL mark IV. Prepare a contrast paper to determine burette volume and titration endpointPre Lab Questions: (Each answer is to be written as a complete sentence) What is the reason for washing the precipitate with water in Step 9? Define precipitate. Define filtrate. In Step 2, what is the purpose for rinsing the stirring rod? read the Procedure to answer the questions Using a balance, mass between 1.50 – 2.00 grams of sodium carbonate in a pre-massed 150mL beaker. Add 20 mL of distilled water and stir thoroughly to make sure all the crystals are dissolved. Rinse the stirring rod with a little distilled water after stirring. Using a balance, mass between 1.50 – 2.00 grams of calcium chloride dihydrate in a pre-massed 50 mL beaker. Repeat Step 2 for the solution in the 50 mL beaker. Pour the calcium chloride solution into the 150mL beaker containing the sodium carbonate solution and stir. Mass a piece of filter paper. Fold the filter paper and place it into the funnel. Wet it with a little distilled water to ensure that it is stuck to the sides of the funnel. Slowly…EXPERIMENT 5; Acid-Base Titration A. Trial 1 Trial 2 Trial 3 5.210 g Mass of KHP 5.210 g 5.210 g (C3H,O4) Mole of KHP Initial Burette 0.00 0.50 5.00 Volume NaOH (mL) Final Burette 31.00 26.25 29.95 Volume NaOH (mL) Total Volume of NaOH (mL) Moles of NaOH Molarity of NaOH (mol/L) Calculate the average Molarity of NaOH B. Titration of Vinegar (HC,H3O2) A 2.352 g sample of Vinegar is titrated with 0.08751M NaOH, and it requires 22.31 mL of NaOH to reach the endpoint. Calculate the mass percent of acetic acid (HC,H;O,) in the vinegar sample? Hint for Calculation of Mass percent of Vinegar (HC,H;O2) 1. Calculate mole of HC,H;O2 from concentration (M) and volume of NaOH (L) needed to titrate Vinegar 2. Calculate mass (g) of Vinegar from mole of Vinegar 3. Mass % Vinegar = (calculated mass Vinegar/ 2.352 g) X 100 C. A solution of malonic acid, H;C;H¿O4 ¸was standardized by titration with 0.100 M NaOH solution. If 20.76 mL of the NaOH solution were required to neutralize completely 13.15 mL of…
- What volume of 0.220 M HBr solution (in mL) is required to obtain 0.0600 moles of HBr? A) 3.67 L B) 273 mL C) 2.73 x 10-4 mL D) 13.2 mL Tap here or pull up for additional resources (1. A solution contains a mixture of Na2CO3 and NaOH. Using phenolphthalein indicator 25 mL of mixture required 19.5 mL of 0.995 M HCl for the endpoint. With methyl orange, 25 mL of solution required 25 mL of the same HCl for the endpoint. Calculate grams per liter of Na2CO3 in the mixture. a.23b.36 c.78 d.112. A 1.2040 g sample containing sodium carbonate and inert material was dissolved in water and titrated to methyl orange end point, requiring 32.50 mL of 0.1020 M HCl. Calculate the %Na2CO4 in the sample. a. 93.36% b. 14.59% c. 12.75% d. 23.41%3. A 0.1510 g KHP (MW = 204.22 g/mol) sample required a volume of 57.04 mL of NaOH solution to reach a phenolphthalein end point. Calculate the molarity of NaOH titrant. a. 0.02593 M b. 0.006481 M c. 0.01296 M d. 0.01012 M1. Molarity of the NaOH solution 0.238 mol/L Trial 1 Trial 2 Trial 2. Volume of H;PO4 added to flask 22.0 mL 22.0 mL 22.0 mL 3. Initial NaOH volume _0.25_ mL _0.75_ mL 0.55 mL 4. Final NaOH volume 17.60_ mL 18.20_ mL _17.90_ mL 5. NAOH volume used for titration to reach green end point mL mL mL 6. NaOH volume used for titration 7. Moles of NaOH used for titration mol mel mel 8. Moles of H;PO4 that reacted mol mel mol 9. Volume of H;PO4 added to flask L 10. Molarity of H3PO4 mol/L mol/L mol/L
- Part A: Standardization of a Sodium Hydroxide Solution Titration 1 Titration 2 Titration 3 Mass of 125 mL flask 45.849g 46.715g 44.953g Mass of flask and KHP 46.849g 47.745g 46.003g Initial buret reading (mL) 0.5 ml 0.5 ml 0.5 ml Final buret reading (mL) 27.8 ml 26.5 ml 26.7 ml Volume of NaOH used (mL) 45.11 ml 45.06 ml 45.14 ml Calculations Titration 1 Titration 2 Titration 3 Moles of KHP Moles of NaOH Molarity of NaOH Average Molarity of NaOH: _______________3 Solid ammonium bromide is slowly added to 75.0 mL of a 0.283 M silver nitrate solution until the concentration of bromide ion is 0.0583 M. The percent of silver ion remaining in solution is %. Submit Answer $ 4 R F % 5 Use the References to access important values if needed for this question. Retry Entire Group 5 more group attempts remaining T G Cengage Learning | Cengage Technical Support ^ 6 MacBook Pro Y H & 7 U J * 00 8 1 ( 9 K O < ) O L P Previous Email Instructor + = Next Save and ExitCalculate the percentage CH3COOH in a sample of vinegar from the following data.Sample = 15.00 g, NaOH used = 43.00 ml ; 0.600 N H2SO4 used for back titration =0.250 ml ; 1.00ml NaOH is equivalent to 0.0315 g H2C2O4. 2H2O.
- A 2.0 mL of acid was added to three experiments whichs volumes are 2mL ,1mL, and 2 mL. if the avaerage Molarity of NaOH is 1.02 x10^-1 1. Find the molarity of the unknown acid for each experiment. 2. find the volumes of the unknown acid for each of the esperiment.Part 1: Preparation of the Primary Citric Acid Standard 1. Mass of Citric Acid: 4.05 g 2. Volume of Citric Acid Solution: 0.75 ml (at the equivalence point) 3. Moles of Citric Acid: (Molar Mass = 192.0 g/mol) 4. Molarity of Citric Acid Solution: Part 2: Titration of the Sodium Hydroxide Solution (Show your work on page 4) 1. Volume of Citric Acid at the Equivalence Point 2. Moles of Citric Acid at the Equivalence Point: 3. Moles of NaOH at the Equivalence Point 3 NaOH + H.CaHsO, → Na:CH.O, + 3 H,O 4. Volume of NaOH 10.0 mL 5. Calculated Molarity of NaOH: Do not use the dilution equation to calculate the molarity of the sodium hydroxide. The dilution equation cannot be used when a reaction is occurringQuantity Your Data 1. Grams of vinegar sample used for your titration 25.000 g 2. Initial Buret Reading of Sodium Hydroxide solution 10.00 mL 3. Final Buret Reading of Sodium Hydroxide solution 28.00 mL 4. Amount of Sodium Hydroxide Solution used to neutralize the vinegar sample 5. Concentration of NaOH in the NaOH solution 0.050 g/mL 6. Grams of NaOH used to neutralize the vinegar 7. Grams of acetic neutralized by the amount of NaOH 8. Percent acetic acid in the vinegar Hints: For #4, The amount of NaOH used is the difference between the starting buret value and the ending buret value. For #6, Once you calculate the amount of sodium hydroxide used, multiple that value by the concentration of NaOH in the NaOH solution. For #7, Multiply the value obtained in number 6 by the number 1.5. Remember, we learned that every 1 gram of NaOH neutralizes 1.5 grams of acetic acid. For #8, Divide the grams of acetic acid by the grams of vinegar sample and multiply this value by…