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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:

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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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**Titration of a Weak Base with a Strong Acid** Consider the titration of 24.0 mL of 0.200 M Na A (a weak base) and 0.100 M HCl. Calculate the pH of the solution after 9.0 mL of HCl 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:** [Your response here]
Transcribed Image Text:**Titration of a Weak Base with a Strong Acid** Consider the titration of 24.0 mL of 0.200 M Na A (a weak base) and 0.100 M HCl. Calculate the pH of the solution after 9.0 mL of HCl 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:** [Your response here]
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:

NaA + H+↔ NaAH+ 

The equilibrium moles can be determined from ICE table as follows:

Chemistry homework question answer, step 2, image 1

Total volume = 24.0mL + 9.0mL = 33.0 mL

Thus, [NaA] = 3.9mmoles/33.0mL 

[NaAH+] = 0.9mmoles/33.0mL 

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