Help me with the rest p Volume of diluted vinegar sampleused in each titration (mL)Initial buret reading (mL)Final buret reading (mL)Volume of NaOH5Volume original vinegar solution 10.00 mLVolume diluted vinegar solution 50.00ml2Dilution FactorMolarity of NaOH1/4.72x103Moles of NaOHergMoles of acetic acid (diluted)Molarity of acetic acid (diluted)Molarity of acetic acid (original)Percent by mass % (m/m)Average % (m/m)VALUECalculations:TRIAL 1TRIAL 26. 16.1TRIAL 310.00m/10.00ml 10.006.113.013.06.920.07.02.88x107¹ 3.26×10 4 3.30x10 +2.88x10 " 3.26x10" 3.30x104

Given: Volume of stock solution (V1) = 10.0 mL Volume of diluted solution (V2) = 50.0 mL molarity of…

Volume original vinegar solution Volume diluted vinegar solution Dilution Factor Molarity of NaOH Volume of diluted vinegar sample used in each titration (mL) Initial buret reading (mL) Final buret reading (mL) Volume of NaOH Moles of NaOH Moles of acetic acid (diluted) Molarity of acetic acid (diluted) Molarity of acetic acid (original) Percent by mass % (m/m) Average % (m/m) Calculations: VALUE 10.00ML 50.00ml 14.78x 103 TRIAL 1 TRIAL 2 6.1 6.1 10.00m/10.00 10.00m² 13.0 20.0 7.0 2.88×10¹ 3.26x1043.30x104 2.88x10 + 3.26x10" 3.30x104 TRIAL 3 6.1 13.0 6.9

Volume original vinegar solution Volume diluted vinegar solution Dilution Factor Molarity of NaOH Volume of diluted vinegar sample used in each titration (mL) Initial buret reading (mL) Final buret reading (mL) Volume of NaOH Moles of NaOH Moles of acetic acid (diluted) Molarity of acetic acid (diluted) Molarity of acetic acid (original) Percent by mass % (m/m) Average % (m/m) Calculations: VALUE 10.00ML 50.00ml 14.78x 103 TRIAL 1 TRIAL 2 6.1 6.1 10.00m/10.00 10.00m² 13.0 20.0 7.0 2.88×10¹ 3.26x1043.30x104 2.88x10 + 3.26x10" 3.30x104 TRIAL 3 6.1 13.0 6.9

Introductory Chemistry For Today

8th Edition

ISBN:9781285644561

Author:Seager

Publisher:Seager

Chapter9: Acids, Bases, And Salts

Section: Chapter Questions

Problem 9.103E

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Similar questions

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.
Consider the nanoscale-level representations for Question 110 of the titration of the aqueous weak acid HX with aqueous NaOH, the titrant. Water molecules and Na+ ions are omitted for clarity. Which diagram corresponds to the situation: After a very small volume of titrant has been added to the initial HX solution? When enough titrant has been added to take the solution just past the equivalence point? Halfway to the equivalence point? At the equivalence point? Nanoscale representations for Question 110.
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A monoprotic organic acid that has a molar mass of 176.1 g/mol is synthesized. Unfortunately, the acid produced is not completely pure. In addition, it is not soluble in water. A chemist weighs a 1.8451-g sample of the impure acid and adds it to 100.0 mL of 0.1050 M NaOH. The acid is soluble in the NaOH solution and reacts to consume most of the NaOH. The amount of excess NaOH is determined by titration: It takes 3.28 mL of 0.0970 M HCl to neutralize the excess NaOH. What is the purity of the original acid, in percent?
Consider the nanoscale-level representations for Question 111 of the titration of the aqueous strong acid HA with aqueous NaOH, the titrant. Water molecules and Na+ ions are omitted for clarity. Which diagram corresponds to the situation: (a) After a very small volume of titrant has been added to the initial HA solution? (b) Halfway to the equivalence point? (c) When enough titrant has been added to take the solution just past the equivalence point? (d) At the equivalence point? Nanoscale representations for Question 111.
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For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%. Is Kc > 1, < 1, or ≈ 1 for a titration reaction? Kc is the product of all reaction products (choose from: Multiplied by, added to, divided by, subtracted from) the product of the reactants, with all concentrations of reactants and products raised to their respective stoichiometric powers. As the titration proceeds, the amount of reactants (choose from: Increases, decreases, remains the same) and the amount of products (choose from: Increases, decreases, remains the same). Because the value of the numerator is (Choose from much greater than, much less than, approximately equal to) the value of the denominator, the value of Kc will be (Choose from much greater than, much less than, approximately equal to)1 for an effective titration reaction.