A 25.0 mL sample of 0.150 M NH3 is titrated with 0.180 M HCI. Kb of NH3 = 1.76 x 10-5 Show your calculations. a) Write the neutralization reaction. b) What volume of 0.180 M HCI is required to reach the equivalence point? c) What is the pH at the equivalence point? d) Explain h

Chemistry: Principles and Reactions
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Chapter14: Equilibria In Acid-base Solutions
Section: Chapter Questions
Problem 46QAP: A 35.00-mL sample of 0.487 M KBrO is titrated with 0.264 M HNO3. (K b BrO =4.0106)(a) Write a...
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A 25.0 mL sample of 0.150 M NH3 is titrated with 0.180 M HCI. Kb of NH3 = 1.76 x 10-5
Show your calculations.
a) Write the neutralization reaction.
b) What volume of 0.180 M HCI is required to reach the equivalence point?
c) What is the pH at the equivalence point?
d) Explain why the pH is not equal to 7 at the equivalence point.
e) What is pH at the midpoint?
f) What is the pH after 30.0 mL of 0.180 M HCI has been added.
g) Draw a reasonable titration curve label the equivalence point and the midpoint.
A
Transcribed Image Text:A 25.0 mL sample of 0.150 M NH3 is titrated with 0.180 M HCI. Kb of NH3 = 1.76 x 10-5 Show your calculations. a) Write the neutralization reaction. b) What volume of 0.180 M HCI is required to reach the equivalence point? c) What is the pH at the equivalence point? d) Explain why the pH is not equal to 7 at the equivalence point. e) What is pH at the midpoint? f) What is the pH after 30.0 mL of 0.180 M HCI has been added. g) Draw a reasonable titration curve label the equivalence point and the midpoint. A
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