b) James performed an acid base titration experiment. A 25 mL sample containing HCl was transferred to Erlenmeyer flask using volumetric pipette. HCl sample was titrated against standardized 0.5010 M NAOH. An average titre volume of 24.35 mL was reported. Suggest a suitable pH indicator for James to be used in this experiment. Predict the initial concentration of the HCl. c) Alan and friends wish to sketch a weak acid strong base titration curve. Before they can do so they need some data points. In a case where 50 mL of 0.20 M CH3COOH (Ka = 1.80 x 10-5 ) was titrated with 0.05 M NaOH, deduce the pH after addition of different volumes of NaOH. (i) 0 mL NaOH

b) James performed an acid base titration experiment. A 25 mL sample containing HCl was transferred to Erlenmeyer flask using volumetric pipette. HCl sample was titrated against standardized 0.5010 M NAOH. An average titre volume of 24.35 mL was reported. Suggest a suitable pH indicator for James to be used in this experiment. Predict the initial concentration of the HCl. c) Alan and friends wish to sketch a weak acid strong base titration curve. Before they can do so they need some data points. In a case where 50 mL of 0.20 M CH3COOH (Ka = 1.80 x 10-5 ) was titrated with 0.05 M NaOH, deduce the pH after addition of different volumes of NaOH. (i) 0 mL NaOH

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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2. In the second titration, the molar mass of an unknown monoprotic acid is calculated. If 0.100 grams of unknown solid are dissolved in water and titrated with 4.68 mL of the NaOH, what is the molar mass of the solid? HINTS: Molar mass is grams / moles. We have grams from the weighed mass, we can calculate moles of acid from the titration. Monoprotic means that it produces 1 H+ per molecule. This means that our moles of NaOH added and moles of acid are the same. You will need to use your concentration of NaOH from the previous question Which is 0.1186
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please see attached image. Im trying to calulate moles of HCl the volume I drew was 5.0 the volume titrated bas 8.15.
Could someone please help me with D-F? I'd greatly appreciate the help. Thank you.
A 100.0 mL sample of 0.10 M NaOH is titrated with 0.20 M HBr. Determine pH after adding 28 mL of the acid. A) 2.36 B) 9.26 C) 11.13 D) 11.64 E) 12.53
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What is the mmole of NaOH?
A 28.1mL sample of 0.108 M HCI is titrated with 0.175 M NAOH. Calculate the pH after the addition of 36.8 mL of NaOH. Hint: Determine the equivalence point so you know where you are in the titration. Report your answer with two places past the decimal point. Type your answer..
Name: Date: Experiment # 11: Acid-Base Titration: Standardization of NaOH Data and results Data table *Remember to include units Trial 1 Trial 2 Trial 3 Trial 4 Mass of KHC8H4O4 (KHP) Initial buret reading (NaOH) Final buret reading (NaOH) Volume of NaOH used Results table Trial 1 Trial 2 Trial 3 Trial 4 Moles of KHP Moles of NaOH Volume of NaOH (L) Molarity of NaOH Average molarity of NaOH Sample calculation: In the space below show complete calculations for moles of KHP and molarity of NaOH using data from TRIAL 2 only. Include formulas and units.
Which of the following statements is TRUE if a pH indicator was NOT added before titration? a. The concentration of the analyte will be lower than the actual. b. Equivalence point could not be determined c. Endpoint can be observed. d. The concentration of the analyte will be higher than the actual.
A 10.0 ml. solution of 0.300 M NH, is titrated with a 0.100 M HCl solution. Calculate the pH after the following additions of the HCI solution: 0.00 ml., 10.0 ml., 30.0 ml., and 40.0 ml.. Round your answer to 3 decimal places. Note: Reference the K, of acids at 25 °C and K, of bases at 25 °C tables for additional information. Part 1 of 4 0.00 ml. Added, pH= Part 2 of 4 10.0 mL Added, pH = 0.. 0.8
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