MAX OF 15 MINUTES A solution contains NaOH (FM = 40), NaHCO3 (FM = 84), and Na,CO3 (FM = 106), either alone or in a permissible combination. Titration of a 50 ml portion to a phenolphthalein end point requires 16.3 mL of 0.100 M HCI. A second 50.0-mL aliquot requires 38.2 mL of the HCI when titrated to a bromocresol green end point. What is the % w/lw composition of the following? Express answers in two decimal places. If the component is not present, write 0.00 in the answer box. NAOH: 96 NazCO; NaHCOz:

MAX OF 15 MINUTES A solution contains NaOH (FM = 40), NaHCO3 (FM = 84), and Na,CO3 (FM = 106), either alone or in a permissible combination. Titration of a 50 ml portion to a phenolphthalein end point requires 16.3 mL of 0.100 M HCI. A second 50.0-mL aliquot requires 38.2 mL of the HCI when titrated to a bromocresol green end point. What is the % w/lw composition of the following? Express answers in two decimal places. If the component is not present, write 0.00 in the answer box. NAOH: 96 NazCO; NaHCOz:

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Qd 56.
In the titration of a volume of 50 mL of a hydrochloric acid solution with Solution density 1.045 g / mL; 29.71 mL of potassium hydroxide with a concentration of 1,963 M were needed, using phenolphthalein as an indicator. Calculate the% m / m of the hydrochloric acid.
10.00mL of 0.10M acetic acid were collected in an Erlenmeyer flask, and then diluted to 100mL with distilled water. After that, two drops of indicator were added. Then, the buret was filled with 0.100M NaOH. The titration was started by slowly adding NaOH to acetic acid. (Note: Ignore the volume of the indicator added, as the volume is tiny compared to 100mL.) What is the pH of the solution in the Erlenmeyer flask after adding 10.00mL of 0.100M NaOH? Question 3 options: 7.22 6.9 8.35 7.00
I did an experiment called Potentiometric Analyses where we titrate NaOH with a solid unknown weak acid. Our group was told that our weak acid is diprotic. Mass of dry, solid acid = .155g Molar concentration of NaOH (mol/L)= 0.100 Buret reading of NaOH, initial (mL)= 50 mL Buret reading NaOH at stocichiometric point, final (mL)= 22.1 mL Volume of NaOH dispensed (mL) 27.9 mL Moles of NaOH to stoichiometric point (mol) : _____________ (I think the above answer would be 2.79 x 10-2? if yes then skip) Moles of acid (mol) _____________ Molar mass of acid (g/mol) __________ Note: In the book it says for a diprotic acid, 2 mol of OH-, neutralizes one mole of acid.
A solution contains both NaHCO3 and Na2CO3. Titration of a 50.00 mL portion to a phenolphthalein end point requires 21.25 mL of 0.1385 M HCl. A second 50.00 mL aliquot requires 45.48 mL of the HCl solution when titrated to a bromocresol green end point. Calculate the molar concentration of Na2CO3 in the original solution. (Do not include the unit for molarity. Enter your answer to three significant figures.)
Part C. Determination of Unknown Balanced chemical equation: Hcl+ Na OH -o Nacl t Hg0 Unknown number NA Trial 1 Trial 2 Trial 3 Volume of unknown HCl titrated 10.00mL NA NA Initial burette reading 0.90mL Final burette reading 30.90mL Volume NaOH added 20 00ML 50.00MLXO.144 m=43.2m moles NaOH added 4.335m (Use average molarity from part B and multiply it by the volume you added) 0.4335 M moles HCl 4.335m 04335m 10.00 mL molarity of HCl Actual Molarity (from instructor) 0.3150M percentage error
The following titration curve shows the titration of triethylamine (C2H5]3N with hydrochloric acid. Which net ionic reaction explains the pH of the solution after 60 mL of titrant is added? 12 10 PH 50 mL of HCI 10 30 70 90 110 O H,O(1) --> H*(aq) + OH (aq) (C,H5)3N(aq) + H*(aq) --> (C,H5)3NH* (aq) (C,H5)3N(aq) + H3O*(aq) (C,H5)3NH*(aq) + OH (aq) (C,H5)3NH (aq) + H2O(1) (C,H5)3N(aq) + H3O*(aq) (C2H5)3N(aq) + H20(1) OH (aq) + (C2H5)3NH*(aq) O O O O O
Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 x 10-5 ), with 0.1000 M NaOH solution after the following additions of titrant. (a) 13.00 mL: pH = (b) 20.40 mL: pH = (с) 30.00 mL: pH =
Paylben
Be sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3CH2)3N (K₁ = 5.2 × 10-4), with 0.1000 M HCl solution after the following additions of titrant. (a) 11.00 mL: pH = (b) 20.10 mL: pH= = (c) 29.00 mL: pH = X X X
Sulfur containing compound weighing 4.8670 grams was digested and purified to form H,SO4. A 25.00-mL of 0.0090 M NaOH was added to the collected H,SO4, and the excess base was then back titrated with 13.45 mL of 0.0100 M HCl. Calculate the sulfur content in the sample in parts per million. This problem involves back titration. Write the two chemical equations involved in the analysis. Identify the analyte, “bridging" reagent and titrant. Solve for the mole of the analyte using the titration values and the stoich relationships. One mole of H2SO4 is equal to one mole of sulfur in the sample. Solve for the mass of S then the ppm S.
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