10111213141516171819202122A chemist titrates 180.0 mL of a 0.4289 M pyridine (C,H,N) solution with 0.5274M HCI solution at 25 °C. Calculate the pH at equivalence. The p K, ofpyridine is 8.77.Round your answer to 2 decimal places.d-Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HCl solution added.PH =%3DSubmit AssignmentContinueMacBook AirF12F10F9F8F7000000* ESF6F4escF2F3F1&*

Given, [C5H5N] = 0.4289 M Volume of C5H5N in mL = 180.0 mL [HCl] = 0.5274 M pKb of C5H5N = 8.77…

A chemist titrates 180.0 mL of a 0.4289M pyridine (CH,N) solution with 0.5274M HCI solution at 25 °C. Calculate the pH at equivalence. The p K, of pyridine is 8.77. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HCl solution added. pH =

A chemist titrates 180.0 mL of a 0.4289M pyridine (CH,N) solution with 0.5274M HCI solution at 25 °C. Calculate the pH at equivalence. The p K, of pyridine is 8.77. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HCl solution added. pH =

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...

See similar textbooks

Similar questions

A chemist titrates 90.0 mL of a 0.4988 M butanoic acid (HC3H,CO₂) solution with 0.0732M KOH solution at 25 °C. Calculate the pH at equivalence. The p K of butanoic acid is 4.82. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added. pH = X S
Please be correct!!
A chemist titrates 130.0 mL of a 0.7778 M lidocaine (C,„H NONH) solution with 0.1602 M HCl solution at 25 °C. Calculate the pH at equivalence. The 14 pK, of lidocaine is 7.94. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HCl solution added. dlo pH = | Explanation Check 2021 McGravw-Hill Education. All Rights Reserved Terms of Use PrivacyI Accessibility • 12:00 acer cspa 08. p
An analytical chemist is titrating 83.7 mL of a 0.3300M solution of isopropylamine ((CH,) CHNH, with a 0.5300M 2 solution of HNO3. The p K, of isopropylamine is 3.33. Calculate the pH of the base solution after the chemist has added 42.4 mL of the HNO, solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO, solution added. Round your answer to 2 decimal places. pH = | ?
An analytical chemist is titrating 179.6 mL of a 0.8400M solution of cyanic acid (HCNO) with a 1.200M solution of KOH. The p K, of cyanic acid is 3.46. Calculate the pH of the acid solution after the chemist has added 144.3 mL of the KOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of KOH solution added. Round your answer to 2 decimal places. pH = |
A chemist titrates 110.0 mL of a 0.1235M butanoic acid (HC,H,CO,) solution with 0.0643 M KOH solution at 25 °C. Calculate the pH at equivalence. The pK, of butanoic acid is 4.82. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added. pH = ] ? %3D
A chemist titrates 60.0 mL of a 0.2562M benzoic acid (HC6H5CO₂) solution with 0.0811M KOH solution at 25 °C. Calculate the pH at equivalence. The pK of benzoic acid is 4.20. a Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added. pH = X S
A chemist titrates 160.0 mL of a 0.4056M hypochlorous acid (HC1O) solution with 0.4300M NaOH solution at 25 °C. Calculate the pH at equivalence. The PK of hypochlorous acid is 7.50. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added. pH = 0 X S
An analytical chemist is titrating 149.7 mL of a 0.5500M solution of benzoic acid (HC H CO,) with a 0.8000 M solution of NaOH. The p K of benzoic acid is 4.20. Calculate the pH of the acid solution after the chemist has added 16.29 mL of the NaOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. Round your answer to 2 decimal places. do pH = Check Explanation 2021 McGraw-Hil Education. All Rights Reserved. Terms of Use Privacy Accessibility V I 11:55 acer &
A chemist titrates 220.0 mL of a 0.6204 M butanoic acid (HC₂H,CO₂) solution with 0.2746 M KOH solution at 25 °C. Calculate the pH at equivalence. The PK of butanoic acid is 4.82. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added. pH = 0 X Ś
Consider the titration of 40.0 mL of 0.200 mol/L HCOONa (aq) with 0.300 mol/L HCI (aq). Ka for HCOOH is 1.8×104. Determine the pH of the original solution. Use the given values to set up an ICE table. Determine the unknown equilibrium concentrations. Initial (M) Change (M) Equilibrium (M) 1 HCOONa(aq) H₂O(1) HCOOH(aq) OH (aq) 5 RESET
A chemist titrates 120.0 mL of a 0.3246M cyanic acid (HCNO) solution with 0.2427M NaOH solution at 25 °C. Calculate the pH at equivalence. The p Ka of cyanic acid is 3.46. Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added. pH = [] X Ś