average molarity of NaOH (from part A), M 0.5507 m NaDH HC,H,0,(aq) + NaOH(aq) -→ H,O(1) + NaC,H,O,(aq) Trial 1 Trial 2 volume of vinegar, mL 10.00 mL 10.00 mL volume of vinegar, L initial buret reading, mL 0.00 mL 0.00mL final buret reading, mL 38.50 mL るい50mL volume of NaOH, mL 38.50mh 36.50mL volume of NaOH, L 0.0383 mL 0.03し5 レ moles of NaOH, mol moles of HC,H,O,(aq), mol (HC,H,0,(aq)], M average [HC,H,O,(aq)], M

average molarity of NaOH (from part A), M 0.5507 m NaDH HC,H,0,(aq) + NaOH(aq) -→ H,O(1) + NaC,H,O,(aq) Trial 1 Trial 2 volume of vinegar, mL 10.00 mL 10.00 mL volume of vinegar, L initial buret reading, mL 0.00 mL 0.00mL final buret reading, mL 38.50 mL るい50mL volume of NaOH, mL 38.50mh 36.50mL volume of NaOH, L 0.0383 mL 0.03し5 レ moles of NaOH, mol moles of HC,H,O,(aq), mol (HC,H,0,(aq)], M average [HC,H,O,(aq)], M

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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Molarity of (NH3) solution (M) from bottle- 5.0 Initial reading of buret (NH3) (mL)- .27 Final reading of buret (NH3) (mL)- 8.25 Volume of Cd(NO3)2 solution (mL)- 10.00 Volume of Na2C2O4 solution (mL)- 10,00 please find Total volume of solution after titration (mL) Total moles of C2O42- (mol) Molarity of C2O42- (M) Total moles of Cd2+ (mol) Moles of [Cd(NH3)4]2+ (mol) Molarity of [Cd(NH3)4]2+ (M) Moles of NH3 added by titration (mol) Moles of NH3 that did not react with Cd2+ (mol) Molarity of NH3 that did not react with Cd2+ (M) Kf for [Cd(NH3)4]2+
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 volumetric flask. A 25.00 mL aliquot from the diluted vinegar required 25.55 mL of 0.1005 M NaOH to reach the phenolphthalein endpoint. Which is the correct equation between the analyte and titrant reaction? CH3COOH + NaOH → NaCH3COO + H2O 2CH3COOH + NaOH → NaCH3COO + H2O C20H14O4 + NaOH → NaC20H14O4 + H2O CH3COOH + 2NaOH → NaCH3COO + H2O
cedric and astrid titrated a 15.00 ml aliquot of grapefruit juice with a 0.134 M NaOH solution to the end point. the initial buret reading was 1.04 ml and the final buret reading was 24.83ml. H3C6H5O7(aq) + 3 NaOH(aq) yeilds Na3C6H5O7(aq) + 3 H2O)(l) The volume of NaOH titrated is 23.79ml 24.83ml - 01.04ml = 23.79ml of NaOH ***What is the mass of citric acid in the juice sample? 0.204g of H3C6H5O7 ( can you please show how to calculate this answer)
Q: The solubility product of Zn(OH)2 at 25oC is 3.0 x 10-6 M3. Calculate the solubility of Zn(OH)2 in water in gdm-3. [Ar: Zn = 65.5; O = 16; H = 1] The Mr for Zn(OH)2 is 99.5. When I multiply it with Ksp, I didn't get the same answer as the given answer which is 0.904 gdm-3.
If you are making 500 ml of a stock solution of 10X PBS pH7.4 please tell me how much of each component you would need to add if the appropriate concentrations of 1x PBS are: 1x PBS Solution Concentrations: NaCl: 137 mM KCl: 2.7 mM Na2HPO4: 10 mM KH2PO4: 1.8 mM MW NaCL= 58.44 g/mol MW KCl = 74.55 g/mol MW Na2HPO4= 141.96 g/mol MW KH2PO4= 136.09 g/mol You should also describe the step by step process of making the solution.
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,…
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
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5 SO2 + 2 MnO4- + H2O ---> 5 SO42- + 2 Mn2+ + 4 H+ The MnO4- solution is poured in a burette and then titrated till the equivalent point is aquired. How many L are needed of MnO4? (the burette starts at 5,7 and is 37,3 after titration)
A 25.0mL mixture solution contains NaNO3 and KIO4, titrated against 0.02M Na2S2O3. The volume of titrants is 18.2 mL, the concentration of KIO4 is:a) 3.64 × 10-4 b) 9.1 × 10-5c) 1.82 × 10-3 d) 2.35e) 3.64 × 10-4
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