Consider the “window cleaner” problem, question 2 on page 12 of your lecture notes. Suppose that prior to the titration, the 10 mL sample of window cleaner was diluted by adding about 10 mL of DI water so that the solution could have better contact with the pH meter. What effect would this dilution have on the volume of titrant needed? (a) The amount of titrant would be the same because the moles of analyte was not changed. (b) The amount of titrant would be doubled because the analyte solution volume was doubled. (c) The amount of titrant would be halved because the was diluted by half.
Consider the “window cleaner” problem, question 2 on page 12 of your lecture notes. Suppose that prior to the titration, the 10 mL sample of window cleaner was diluted by adding about 10 mL of DI water so that the solution could have better contact with the pH meter. What effect would this dilution have on the volume of titrant needed? (a) The amount of titrant would be the same because the moles of analyte was not changed. (b) The amount of titrant would be doubled because the analyte solution volume was doubled. (c) The amount of titrant would be halved because the was diluted by half.
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter16: Reactions Between Acids And Bases
Section: Chapter Questions
Problem 16.103QE: A bottle of concentrated hydroiodic acid is 57% HI by weight and has a density of 1.70 g/mL. A...
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Consider the “window cleaner” problem, question 2 on page 12 of your lecture notes. Suppose that prior to the titration, the 10 mL sample of window cleaner was diluted by adding about 10 mL of DI water so that the solution could have better contact with the pH meter. What effect would this dilution have on the volume of titrant needed?
(a) The amount of titrant would be the same because the moles of analyte was not changed.
(b) The amount of titrant would be doubled because the analyte solution volume was doubled.
(c) The amount of titrant would be halved because the was diluted by half.
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