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A chemistry graduate student is given 250. mL of a 0.40M pyridine (C₂H₂N) solution. Pyridine is a weak base with K₁=1.7 × 10⁹. What mass of C₂H₂NHBr should the student dissolve in the C₂H₂N solution to turn it into a buffer with pH = 4.75? You may assume that the volume of the solution doesn't change when the CH NHBr is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.

A chemistry graduate student is given 250. mL of a 0.40M pyridine (C₂H₂N) solution. Pyridine is a weak base with K₁=1.7 × 10⁹. What mass of C₂H₂NHBr should the student dissolve in the C₂H₂N solution to turn it into a buffer with pH = 4.75? You may assume that the volume of the solution doesn't change when the CH NHBr is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.

Chemistry
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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A chemistry graduate student is given 250. mL of a 0.40M pyridine (C₂H₂N) solution. Pyridine is a weak base with K=1.7 x 10. What mass of C H₂NHBr should the student dissolve in the C,H-N solution to turn it into a buffer with pH =4.75? You may assume that the volume of the solution doesn't change when the C₂H₂NHBr is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.
Transcribed Image Text:A chemistry graduate student is given 250. mL of a 0.40M pyridine (C₂H₂N) solution. Pyridine is a weak base with K=1.7 x 10. What mass of C H₂NHBr should the student dissolve in the C,H-N solution to turn it into a buffer with pH =4.75? You may assume that the volume of the solution doesn't change when the C₂H₂NHBr is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.
Expert Solution
Step 1: Interpretation of datas :

Given ,

V o l u m e space o f space p y r i d i n e space equals space 250. space m L M o l a r i t y space o f space p y r i d i n e space equals space 0.40 space M K subscript b equals 1.7 cross times 10 to the power of negative 9 end exponent p H equals space 4.75 o b j e c t i v e space colon C a l c u l a t e space m a s s space o f space C subscript 5 H subscript 5 N H B r space.

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