Macmillan Learning Question 7 of 20 > Attempt 1 To a 25.00 mL volumetric flask, a lab technician adds a 0.125 g sample of a weak monoprotic acid, HA, and dilutes to the mark with distilled water. The technician then titrates this weak acid solution with 0.0866 M KOH. She reaches the endpoint after adding 47.47 mL of the KOH solution. Determine the number of moles of the weak acid in the solution. moles of weak acid: 0.00369441 Incorrect Answer Determine the molar mass of the weak acid. 33.8359 molar mass= Incorrect Answer mol g/mol After the technician adds 17.39 ml. of the KOH solution, the pH of the mixture is 5.33. Determine the pKa of the weak acid. 4.68858 PK₁ = Incorrect Answer

Macmillan Learning Question 7 of 20 > Attempt 1 To a 25.00 mL volumetric flask, a lab technician adds a 0.125 g sample of a weak monoprotic acid, HA, and dilutes to the mark with distilled water. The technician then titrates this weak acid solution with 0.0866 M KOH. She reaches the endpoint after adding 47.47 mL of the KOH solution. Determine the number of moles of the weak acid in the solution. moles of weak acid: 0.00369441 Incorrect Answer Determine the molar mass of the weak acid. 33.8359 molar mass= Incorrect Answer mol g/mol After the technician adds 17.39 ml. of the KOH solution, the pH of the mixture is 5.33. Determine the pKa of the weak acid. 4.68858 PK₁ = Incorrect Answer

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter31: Introduction To Analytical Separations

Section: Chapter Questions

Problem 31.17QAP

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A diprotic acid, H2B(MM=126g/moL), is determined to be a hydrate, H2B xH2O. A 10.00-g sample of this hydrate is dissolved in enough water to make 150.0 mL of solution. Twenty-five milliliters of this solution requires 48.5 mL of 0.425 M NaOH to reach the equivalence point. What is x?
A bottle of concentrated hydroiodic acid is 57% HI by weight and has a density of 1.70 g/mL. A solution of this strong and corrosive acid is made by adding exactly 10.0 mL to some water and diluting to 250.0 mL. If the information on the label is correct, what volume of 0.988 M NaOH is needed to neutralize the HI solution? Suggest an indicator for the titration.
8-115 When a solution prepared by dissolving 0.125 g of an unknown diprotic acid in 25.0 mL of water is titrated with 0.200 M NaOH, 30.0 ml, of the NaOH solution is needed to neutralize the acid. Determine the molarity of the acid solution. What is the molar mass of the unknown diprotic acid?
8-47 A sample of 27.0 mL of 0.310 M NaOH is titrated with 0.740 M H2SO4. How many milliliters of the H2SO4 solution are required to reach the end point?
What is an acidbase indicator? Define the equivalence (stoichiometric) point and the end point of a titration. Why should you choose an indicator so that the two points coincide? Do the pH values of the two points have to be within 0.01 pH unit of each other? Explain.
8-46 If 25.0 mL of an aqueous solution of H2SO4 requires 19.7 mL of 0.72 M NaOH to reach the end point, what is the molarity of the HO2SO4 solution?
Give typed full explanation not a single word hand written otherwise leave it
Question 18 of 30 Submit A 20.0 mL solution of NaOH is neutralized with 26.3 mL of 0.200 M HBr. What is the concentration of the original NaOH solution? | M 1 4 6. C 7 +/- x 10 0 Tap here or pull up for additional resources 3. LO 00
One-half liter of 2.7M HI is mixed with 200 ml of 1.61 M of H2SO4. What is the concentration of H+ in the resulting solution? Enter a number to 4 decimal places and in M of H+.
100 mL of 0.10 M NH3 (Kb = 1.8 x 10-5) are mixed with 100 mL of 0.10 M HCN (Ka = 4.9 × 10-10). Which statement is true? (3) The resulting solution is neutral O (2) The resulting solution is basic O (4) No answer is correct as not enough information is given. O (1) The resulting solution is acidic E D Moving to the next question prevents changes to this answer. 4 R F % S T G 6. & 7 H C 8 00 K M Question 2 of 37
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1 Settings „l ? 9:57 AM • 97% kzb-vvdv-bir » Grace Dykstra is presenting TAM Mo Fa 2 H the amount of water added was 1000 mL instead. Would t V ז ןיידצל 1oODmL - LoNIcediud Lies solvent) concentrated than the initial KOOL-AID? Examples: Concentration Calculations Part A Determining Concentration from moles and V: (answers: 0.46 mel; 247 mol ; 0.658 ma) 1. Determine the molar concentration when 0.23 mol of sodium sulfate are dissolved in 500ml of water. G: lo00 0.50ol 0.a3 mol 0.50OL 0. a3 mol = 0.4b mol /L mol 0.500 L 1. 2. Determine the concentration when 1.20 moles of sodium hydroxide dissolves into 450 ml of water. 3. Determine the concentration of a Nacl solution if 25.0 g of sodium chloride is dissolved into 650 ml of water. Alyssa Khadija You Grac 22 others