An ecobotanist separates the components of a tropical barkextract by chromatography. She discovers a large proportion ofquinidine, a dextrorotatory isomer of quinine used for control of arrhythmic heartbeat. Quinidine has two basic nitrogens (K(b1)=4.0X10⁻⁶ and K(b2)=1.0X10⁻¹⁰). To measure theconcentration, she carries out a titration. Because of the low solubility of quinidine, she first protonates both nitrogens withexcess HCl and titrates the acidified solution with standardized base. A 33.85-mg sample of quinidine (M=324.41g/mol) isacidified with 6.55 mL of 0.150 MHCl.(a) How many milliliters of 0.0133 MNaOH are needed to titrate the excess HCl?(b) How many additional milliliters of titrant are needed to reach the first equivalence point of quinidine dihydrochloride?(c) What is the pH at the first equivalence point?

An ecobotanist separates the components of a tropical barkextract by chromatography. She discovers a large proportion ofquinidine, a dextrorotatory isomer of quinine used for control of arrhythmic heartbeat. Quinidine has two basic nitrogens (K(b1)=4.0X10⁻⁶ and K(b2)=1.0X10⁻¹⁰). To measure theconcentration, she carries out a titration. Because of the low solubility of quinidine, she first protonates both nitrogens withexcess HCl and titrates the acidified solution with standardized base. A 33.85-mg sample of quinidine (M=324.41g/mol) isacidified with 6.55 mL of 0.150 MHCl.(a) How many milliliters of 0.0133 MNaOH are needed to titrate the excess HCl?(b) How many additional milliliters of titrant are needed to reach the first equivalence point of quinidine dihydrochloride?(c) What is the pH at the first equivalence point?

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.20QAP

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an ecobotanist separates the components of a tropical barkextract by chromatography. She discovers a large proportion ofquinidine, a dextrorotatory isomer of quinine used for control of arrhythmic heartbeat. Quinidine has two basic nitrogens (K(b1)=4.0X10⁻⁶ and K(b2)=1.0X10⁻¹⁰). To measure theconcentration, she carries out a titration. Because of the low solubility of quinidine, she first protonates both nitrogens withexcess HCl and titrates the acidified solution with standardized base. A 33.85-mg sample of quinidine (M=324.41g/mol) isacidified with 6.55 mL of 0.150 MHCl.(c) What is the pH at the first equivalence point?
A commercial vinegar was analyzed by titration to determine the percent acetic acid. Briefly, 10.00 mL of vinegar sample was diluted to 100. mL solution in a volumetric flask. A 25.00 mL aliquot from the diluted vinegar required 25.55 mL of 0.1005 M sodium hydroxide to reach the phenolphthalein endpoint From this experiment which is the standard solution used in titration? CH3COOH NaOH NaCH3COO C20H14O4
1 ) The density of a 5.26MNaHCO 3 (84.0 g/mol) is 1.19g / m * l . Its molality is 2) Calculate the pAg^ + at the equivalence point in the titration of 25.0ml of 0.0823 M Kl with 0.051M AgNO 3 . Ksp Agl=8.3*10^ -16 3) Commercial concentrated aqueous nitric acid is 70.4% HNO3(63.0 g/mol) by mass and has a density of 1.41g / m * l . The molarity of this solution is 4) Consider the titration of 25ml of 0.0823M KI with 0.051M AGNO3, Kspagi =8.3x10-16 Calculate pAg* after adding 39.0 ml I03 Ag* + 103 ====AglO3 5) Commercial concentrated aqueous nitric acid is 70.4 1\%HNO 3 (63.0 g/mol) by mass and has a density of 1.41g / m * l . The molarity of this solution is: 6) What mass in g of Na 2 CO 3 [106 g/mol] is required to prepare 250 ml of 0.3M aqueous solution in Na^ + [23.g/mol] ? 7) Calculate pAg^ + after adding 42.30ml AgNO 3 In the titration of of 0.0823M Kl with 0.051M AgNO 3 . KspAgl = 8.3 * 10 ^ - 16 8) The milliliters of concentrated HClO 4 (100.5 g/mol),6 60% by mass,…
1. An instrument used to measure the acidity or alkalinity of a solution. 2. Chemicals of high purity which are used to standardize a solution 3. Used as standard but have a tendency to lose water from crystallization.
Ammonium acetate buffer is used as an important reagent for studying molecular biology, biological buffers, reagents and DNA and RNA purification. Ammonium acetate (0.24M) in combination with cold 66% (v/v) ethanol quantitatively precipitated RNA from very dilute solutions (greater than or equal to μg/ml) after centrifugation. Ammonium acetate (CH3COONH4) is also a salt of weak acid (CH3COOH) and weak base (NH4OH). The Ka and Kb are equal to 1.8 x10-5. The pH of this salt solution will be, a. pH > 7 b. pH ≈ 7 c. pH < 7 d. both cation and anion don’t hydrolyze. e. none of the above.
1ml of 1mg/ml of diluted aspirin powder solution is titrated by 0.005M NaOH. Before titration, 5 ml ethanol, 14 ml CO2 free water, and 4 drops bromothymol blue indicator is added into diluted aspirin powder solution. Mass of 0.005M NaOH used is obtained by weight titration = 1.1384g . [Molar mass of aspirin =180.15]; for solution, 1g=1mL at 25 degree Celcius. Calculate the purity of the powder.
A reprecipitation was employed to remove occluded nitrate from BaSO4 precipitate prior to isotopic analysis of oxygen for geologic studies.9 Approximately 30 mg of BaSO4 crystals were mixed with 15 mL of 0.05 M DTPA in 1 M NaOH. After dissolving the solid with vigorous shaking at 708C, it was reprecipitated by adding 10 M HCl dropwise to obtain pH 3–4 and allowing the mixture to stand for 1 h. The solid was isolated by centrifugation, removal of the mother liquor, and resuspension in deionized water. Centrifugation and washing was repeated a secondtime to reduce the molar ratio NO-3 /SO-4 2 from 0.25 in the original precipitate to 0.001 in the purified material. What will be the predominant species of sulfate and DTPA at pH 14 and pH 3? Explain why BaSO4 dissolves in DTPA in 1 M NaOH and then reprecipitates when the pH is lowered to 3–4.
Mr. Clean recently bought a laboratory-grade sodium carbonate from a chemical company known as Brand X. He was supposed to use it in the production of detergents. Unfortunately, he was scammed by the company. He suspected that he purchased a crude sodium carbonate so he tasked the Quality Assurance Department to determine the components of the purchased chemical. The chemist assigned to analyze the sample used double indicator method. For the standardization of HCl titrant, 0.1025 g Na2CO3 of 99.5% purity (FW: 106.00) required 8.20 mL of the titrant to reach the phenolphthalein endpoint. FW: NaOH (40.00), NaHCO3 (84.01), Na2CO3 (106.00) a. What is the molarity of the titrant? The chemist obtained a 3.150 g sample and dissolved it in distilled water to produce a 50.0 mL solution. An aliquot of 10.00 mL was obtained and diluted in a 100.0 mL volumetric flask. A 50.00-mL aliquot of the diluted sample was taken and it required 25.70 mL of titrant for the methyl orange endpoint, while…
A mixture of 500.0 mg acid A and 250.0 mg acid B is titrated with 0.1150 N NaOH. The endpoint is overstepped on the addition of 45.30 mL of the base. The solution is backtitrated using 1.40 mL of 0.1234 N HCl solution. If the equivalent weight of A is thrice the equivalent mass of B, calculate the equivalent weights of A and B.
A metal M forms a water-soluble hydroxide with a chemical formula of MOH. To determine whatM is, a student prepared 250.0 cm3 of MOH standard solution by dissolving 1.17 g of MOH in distilled water. Then the student titrated 25.0 cm3of the solution with 0.055 M H2SO4(aq) usingphenolphthalein as indicator. The titration was repeated several times and the mean titre was 18.85cm3.(a) Describe how the 250.0 cm3 of MOH standard solution was prepared. (b) (i) Calculate the molar mass of MOH. (ii) Determine what M is. (Relative atomic masses: H = 1.0, O = 16.0) THANKYOU!!!
Part 1: Preparation of the Primary Citric Acid Standard Mass of Citric Acid: 4.05 g Volume of Citric Acid Solution: 7.5 (at the equivalence point) Moles of Citric Acid: 0211 mol (Molar Mass = 192.0 g/mol) Molarity of Citric Acid Solution: 0.2813 help me with part Part 2: Titration of the Sodium Hydroxide Solution Volume of Citric Acid at the Equivalence Point _____________ Moles of Citric Acid at the Equivalence Point _______________ Moles of NaOH at the Equivalence Point __________________ 3 NaOH + H3C6H5O7 à Na3C6H5O7 + 3 H2O Volume of NaOH _______________________________ 5. Calculated Molarity of NaOH _______________________
. What is the change in solubility from 35oC to 55oC?