Consider the titration of 24.0 mL of 0.200 M Na A (a weak base) and 0.100 MH CI. Calculate the pH of the solution after 9.0 mL of H CI has been added. The Kb of the base is 1.93E-5 To avoid rounding errors report your answer with two decimal places. Your Answer:
Consider the titration of 24.0 mL of 0.200 M Na A (a weak base) and 0.100 MH CI. Calculate the pH of the solution after 9.0 mL of H CI has been added. The Kb of the base is 1.93E-5 To avoid rounding errors report your answer with two decimal places. Your Answer:
Chemistry & Chemical Reactivity
10th Edition
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Chapter17: Principles Of Chemical Reactivity: Other Aspects Of Aqueous Equilibria
Section17.3: Acid-base Titrations
Problem 17.6CYU: The titration of 0.100 M acetic acid with 0.100 M NaOH is described in the text. What is the pH of...
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Transcribed Image Text:Consider the titration of 24.0 mL of 0.200 M Na A (a weak base) and 0.100 MH CI.
Calculate the pH of the solution after 9.0 mL of H CI has been added. The Kb of the
base is 1.93E-5
To avoid rounding errors report your answer with two decimal places.
Your Answer:
Expert Solution
Step 1
No of moles of weak base, NaA = Molarity × Volume = 0.200M×24.0mL = 4.80 mmoles
No of moles of strong acid, HCl = Molarity × Volume = 0.100M×9.0mL = 0.90 mmoles
Step 2
When NaA and HCl are titrated, the net ionic equation is:
The equilibrium moles can be determined from ICE table as follows:
Total volume = 24.0mL + 9.0mL = 33.0 mL
Thus, [NaA] = 3.9mmoles/33.0mL
[NaAH+] = 0.9mmoles/33.0mL
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